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Author's Note _____________
The name EUREKA is taken from a charming bit of bathematics. More than 2000 years ago, the King of Greece asked the famous mathematician Archimedes to determine if his crown were pure gold or alloyed with silver. While lounging in the tub, Archimedes realized that a crown of pure gold would displace more water than an equal volume of lighter alloyed gold. So enthralled was he by this discovery that he ran naked through the streets of Syracuse, shouting, "Eureka! I've got it!" Well, I hope you receive the same pleasure from reading this book that Archimedes got from solving his problem or that I got from writing it. You might even want to read it in the bathtub. My thanks go to my sister Emily, who contributed the subtitle. I would also like to thank my agent Betty Marks and editor Paul Heacock. I am also indebted to Miriam Smith and Mary Secor and the other mem bers of the Onteora staff who were instrumental in the creation and writing of Eureka!
7
Errata __________________________________
1. There are no mistakes in this book. The above should read "ERRATUM."
Is the correction a mistake or not?
Contents
1l~ Fun and Games 1. 2. 3. 4. 5.
Twisted Topology A Bag of Tricks and Treats A Score of Games The Magic's There Rubiquity
15 26 36 48
60
~~ Nifty Numerics 6. Palindromesemordnilap 7. A Pole Vaulter 8. A Timely Switch
69 71
73
3)~ Fallacies and Logic 9. If This Is Not a Chapter, My Name Is Raymond Smullyan 10. Thrice Befuddled 11. Better MixedUp Than Lost
75 89 92 11
~g And Even Dissection of Solids 12. Archimedes Anderson and the Case of the Sinister Plot 13. How to Dissect a Square and Other Marvels of Modern Biology 14. Geometer's Heaven 15. Hole in the Sphere 16. Convexstasy 17. Great Unsolved Problems 18. Out of This World
93 96 105 108 110 113 115
@g Photons Are Light Matter, Too 19. Archimedes Anderson and the Gambling Candidate 20. Once Upon a Time. . . 21. A Problem Fly 22. The Leading Series of Pisa 23. The Early Something Catches the Whatever
119 122 123 124 127
®g Shortcuts 24. A Speedier System of Solving 25. A Letter Home 26. In Which We Are Initiated Into the Secret Society of Square Root Solvers 27. Heads and Legs 28. Noble Bases 29. A Division in Ancient Rye 30. Getting at the Root of the Problem 31. Or is it 32? Remumbt:r Nembers 12
EUREKA!
129 135 137 141 143 145 146 149
'1h
Neat Numbers
32. Prime Time 33. A Sense of Balance 34. Perfect Numbers and Some NotSoPerfect Numbers
151 159 161
f8)~ Cranium Crackers and Cheese: Problems to Munch On 35. Classy Problems 36. LEITERS + DIGITS
=
FRUSTRATION
165 174
~~ FUNdamental Ratios 37. Expand Your Mind 38. E? Ah! 39. A Section of Gold 40. A BundledUp Buyer 41. A Piece of Pi Bibliography
181 183 188 195 196 202
13
110 Fun and Games 1.
Twisted Topology _ _ _ _ _ _ _ __
A limerick writer once noted: A mathematician confided That a Mobius band is onesided. And you'll get quite a laugh If you cut one in half, For it stays in one piece when divided! The Mobius band referred to is one of the most curious shapes in all topology. Introduced by one August Ferdinand Mobius, a German mathematician, the Mobius band or Mobius strip has been described as a strip which has no "other side."
15
While this oddity is difficult to imagine, it is surprisingly easy to construct. Merely cut out a strip of paper at least one inch wide and eight inches long. Mark one long edge with an A in the left corner and a B in the right corner. Then mark the parallel edge B in the left corner and A in the right corner. Tape the short edges together so that one A is above the other A, and one B is over the other B. You've made a ring. This may not seem like much, but the best is yet to come. Now make a mark on one side of the paper. (It doesn't matter which side, as you'll see in a moment.) Using a finger, trace around the ring. Soon your finger and your mark will be on opposite sides of the paper, and soon after that your finger will be right on top of the mark! You have just proved that the Mobius strip has one and only one side (by showing the "two" sides are actually joined). "Balderdash!" you say. "Everyone knows that a piece of paper has two sides." Not the Mobius strip. In a mathematician's terms, the band has only one side and one edge. (The Goodrich company has patented the use of the Mobius strip as a conveyor belt; both sides being one, it lasts twice as long as conventional ones.) You may think that you see two of each, but I can assure you that you really don't; they are one and the same. Even odder is the double Mobius strip, formed by putting two strips of paper together, one on top of the other, giving both a halftwist, and joining the ends. If you now make a mark on the inside of the outside band and begin circling the two, you will find that the mark has jumped to the opposite band! A bug crawling between the bands would see your mark alternately on the ceiling and the floor; it would need considerable imagination to comprehend that the floor and the ceiling are actually on one 16
EUREKA!
strip. What appears to be two nested strips is actually one large one! Even more unusual is another property of the Mobius band mentioned in the opening limerick. If you cut a Mobius band down its "middle" (it really has none to speak of), you will not get two Mobius strips but a single, large, twosided strip. Odder is cutting one in thirds; produced is a large, twosided strip intertwined with a new, smaller Mobius strip. Try, for instance, cutting a Mobius band in fifths. The results are guaranteed to be surprising!
QUICKIE
1. If Fig. 11 is cut along the middle, what will the result be? Will the loop(s) be linked or not? See the Answers section at the end of the Chapter.
Fig. 1·1
If you were to join two Mobius strips that are mirror images of each other, you would get another unusual topological oddity, the Klein bottle. Named after one Felix Klein, another German mathematician, it too has
17
inspired a limerick: "A mathematician named Klein/ Thought the Mobius band was divine.! Said he, 'If you glue/ The edges of two,! You'll get a weird bottle like mine.' " The Klein bottle, like the onesided Mobius band, has only one surface and precisely zero edges! It also has no inside! To construct a Klein bottle, bend the end of a tube of material and pass it through its own wall. Then join the two ends, forming a continuous closed surface (Fig. 12). Voihl!
Fig. 12
Normally, the wall of a threedimensional object must be penetrated if one is to reach the other "side." Not so with the Klein bottle. Merely by entering the opening and staying on the "outside," the "inside" is reached. This 18
EUREKA!
means that the outside is the inside, proving that it has one surface only. Are there any practical applications of the Klein bottle? So far there are none; liquids would spill from it rather easily. But if a cap is put over the opening, the bottle serves as its own handle! This area is one that should certainly be explored. The problem that started topology all began in the small Russian town of Konigsberg, a quaint little village cut into four parts by the river Pregel. In summer, the townsfolk liked to take their evening strolls across the seven bridges (Fig. 13).
,~

Fig. 13
Much to their surprise, though, they discovered that they could not cross all the bridges once in a single stroll without crossing a bridge twice or retracing their steps. Try it yourself and see. By the time the problem reached the ears of Leonhard Euler, the great 18th century Swiss mathematician, they had ruined it by adding an eighth bridge and changing the town's name to Kaliningrad. Nevertheless, he drew the basic network by cutting out everything shown in
19
Fig. 13, except the actual routes. If he had drawn the network in one stroke, it would have been equivalent to strolling on the bridges; however, Euler found that the network was not a onestroker, so a sevenbridge walk was impossible. He counted the number of lines leading into each dot representing each part of the town, and called the dots odd if the number was odd and even if the number was even. His final conclusion? A network is a onestroker if all the dots are even or if only two of the dots are odd (if so, the stroll or stroke must begin on an odd dot). 2. Are Figs. 14 and 15 onestrokers? Don't actually draw them!
Fig. 1·4
Fig. 1·5
3. Of a slightly different nature is the topological puzzle in Fig. 16. The problem is to draw a continuous line so that each of the 16 segments is crossed once and only once, without drawing through the vertices or along 20
EUREKA!
the line segments. Can it be done on either a plane, a sphere, or a torus (a bagelshaped solid)?
Fig. 1·6
4. A wellknown brewery uses as its trademark three linked rings (Fig. 17)or are they linked at all? If one single ring is cut, the set falls apart. If you were the King's armorer, could you make him an entire tunic of chain mail using this pattern? How will you place the next ring?
Fig. 1·7
21
At the other end of the spectrum of recreational math are problems of a class of deceptively simple cranium crackers. (An example: I know a math teacher who can remove his vest without first taking off his coat! Of course, he does look a little funny when he does it in restaurants .... ) 5. Can you separate the cup and the string in Fig. 18 without cutting the string, undoing the knot, or breaking the handle? Think!
Fig.l.8
6. This last one I find absolutely perfect at parties and as a toolofthetrade for would·be matchmakers. It's also good for budding escape artists. Loosely tie two sets of wrists as shown in Fig. 19. If the participants don't cut or unknot the ropes (amputation is out of the question), how in the world can they get free?
22
EUREKA!
Fig. 1·9
Answers 1. The result will be two unlinked loops like the first, except for the fact that one is the mirror image of the other. Because of the crossing of the original, the two halves have opposite twists (this is also why they cannot be linked). 2. The first has two odd vertices, so it can be drawn by starting at one of the bottom corners. The second has four odd vertices, and so is not a one·stroker. 3. A line that enters and leaves a space must cross two line segments. Spaces A, B, and C in Fig. 1·10 are each surrounded by an odd number of segments (five), so it follows that the ends of the line segment must be in each; clearly, as a line has only two ends, this is impossible.
Fig. 1·10
23
The same reasoning applies to a sphere and the side of a toruswith one exception that makes the problem solvable. If the hole is situated inside either A, B, or C, the puzzle is reduced to mere linedrawing, as in Fig. 111.
Fig. 111
4. The arrangement is shown in Fig. 112. The pattern can be continued indefinitely.
24
EUREKA!
5. For some odd reason, this one seems to baffle people. Just take the center loop of the string and pull it back until you have a reasonable amount of slack. Then pull it forward around the side of the cup, and let go. 6. It looks hard, but as illustrated in Fig. 113, release is possible by pulling the center of the rope over one wrist and through the loop.
Fig. 113
25
2.
A Bag of Tricks and Treats _ _ _ __
Many numbers and operations of mathematics have special properties that enable them to be easily adapted for mathematical "tricks" or stunts. Most everyone has heard of something akin to the following: Think of a number and mUltiply it by 3. Now add 6, divide by 3, and subtract your first number. What is the result? Chances are that you vaguely suspect how the stunt works. Indeed, it seems the explanation would be lengthy, but it's really only a case of cancellation of fractions. The instructions create the statement (3N + 6)/3  N, where N is the number you thought of. This reduces to (N + 2)  N, or more simply, 2 in all cases. 1. How about this one? Think of a number and mUltiply it by 4. Then add a second, even number and divide the sum by 2. Now subtract half of your second number and again divide by 2. The result is always your first number! How does this work? 2. One clever twist on the "think of a number" trick works extremely well with large numbers of people. At first it looks easy, but it's really very hard to figure out, and even starts to seem impossible! Collar a halfdozen friends and coax each into picking a number between 51 and 100. You yourselfthe prestidigitato~elect a number from 1 to 50 and seal that number in a convenient hat or envelope (or make a copy of the number for each friend). Subtract your selected number from 99 and say 26
EUREKA!
the result aloud; then tell your audience to add that number to each of their selected numbers, cross out the first digit, add that digit to the result, and subtract the answer from the original number. The result is always identical to your sealed number! Can you figure out how this works? Another puzzler that works especially well with large crowds is based on the factorization of a certain number. Find four people, say, AI, Beth, Cathy, and Dan. Ask Al to select any threedigit number and write it twice to form a sixdigit number. Tell him to pass this number to Beth, and ask her to divide this number by 7 (actually, as you can use the successive primes 7, 11, and 13 in any order, you can ask Beth to select a prime between 6 and 16 and then ask Cathy and Dan to each select different primes in the same range). "Don't worry about the remainder," you say. "There won't be any." The quotient is passed to Cathy, who is cordially requested to divide the number by 11, and the new quotient is divided by 13 by Dan, who passes the number back to AI. "Look at that," you say, "and you will find your original number!" Again you are correct! This trick is based on the product of 7, 11, and 13, the number 1001. Writing a threedigit number twice is the same as multiplying it by 1001. Dividing by the factors of 1001 yields the original threedigit number. 3. If you are older than 9 and younger than 100, write your age thrice to form a sixdigit number. Can you prove that this number is divisible by 7 and 13? Is the same true if the number is written only twice? One other favorite is asking a friend to select a digit from 1 to 6. Multiplying this digit by 142,857 results in the same digits in cyclic order! Or, you can ask her to 27
select a digit from 1 to 9, multiply that digit by 9, and multiply the product by 12,345,679; you can also ask her to multiply the digit by 7 and then by 15,873. The results should surprise you! You can impress your peers with your incredible ability as a lightning calculator with this one. Ask the nearest person to write down an excessively large number. In seconds you can tell him if it's divisible by 11 or not. The method? Sum up the digits in the evennumbered places and in the odd. If the difference between the two sums is o or any mUltiple of 11, the number is a multiple of 11. 4. Here's another way. Ask a buddy to write down two numbers and write their sum below them. Then ask her to write the sum of the last two numbers below that, and so on, until there are ten numbers, each the sum of the previous two. Then announce the sum. For example: 2
5 7 12 19 31 50 81 131 + 212
Merely by glancing at the list, you can tell her the total is 550. How? There's a simple rule. 5. A reputation as a mystic can be strengthened by these magical experiments with dice. With your back turned, ask your subject to set up three dice and add the top faces. He 28
EUREKA!
then chooses one die, adds in the number on the bottom, and then adds in the number shown on the top, remembering the grand total of the five faces. Hint: The magician says the total in Fig. 21 is 18. How does he know?
· U •
/
•
•
I
•
•
I
•
I I
/
•
I •
... ...
/ I
•
I
·V
Fig. 2·1
6. Here's a harder trick using the same principle. Ask your subject to stack three dice in a pile and add up the spots on the hidden faces (this means "not the top face," but don't say this!). The sum in Fig. 21 is 16. How do you figure it out?
/
...../ • • • • / • • V I
I
I
II
I
• • • • • V I
I
Fig. 2·2
7. Ask the nearest mathematician about this one: Tell your friend or enemy to arrange two dice right next to each other to form a twodigit number (reading from left 29
to right). Then ask her to form a similar twodigit number by flipping both dice over .
•• •
•
• •
• • • ••
• • • •
•
Fig. 2·3
In Fig. 23, the two numbers are 35 and 42. Request now that she join the two to form a fourdigit number (as in 3542), divide by 11, and tell you the result. You can tell her the original arrangement of the dice by subtracting 7 and dividing by 9. Can you explain how this works? 8. Some stunts just take a little thought to unravel. If someone says to you, "I bet you a dollar that if you give me two dollars I'll give you four in exchange," would that be a good bet to take? 9. Try this one on the nearest unwitting stooge. Ask him for a dollar. Then put it, along with a crisp, new, irresistible bill of your own, into an empty matchbox. Say the appropriate mumbojumbo and reluctantly offer the box and its contents for $1.50. Would you yourself fall for it? Un losable bets like these abound. Have someone place a dollar bill on the table and cover it with any card drawn from a deck (assume for the moment that it's the ace of spades). Now bet the "sucker" 50 cents that he will not answer the value of the card to each of your next three questions. Your first two questions should be totally irrel30
EUREKA!
evant; but make sure you receive two answers of, "the ace of spades." Now ask him what he will take for the bill below the card! Either way he loses four bits. 10. Others are somewhat more mathematical. These deal with properties of even and odd numbers. Coerce a handy millionaire into this wager: You and she will each put down a single die, separately. If the total of both dice and her number are both odd or both even, she wins. Otherwise, you take the pot. What's the catch? Take a full deck of cards and ask a nimble card shark to cut the deck into seven piles. Bet that there will be an odd number of piles containing an even number of cards. You win! Now you can take away or add a card and adjust the bet so that there will be an odd number of piles containing an odd number of cards. Again you win! DID YOU KNOW THAT ... Eight perfect shuffles, in which the cards from each half of the deck wind up in alternating order in the shuffled pack, restores the deck? 11. A real mathematician named Walter Penney came up with the following paradoxical bet. Given the situation where you are tossing a coin three times in succession, it is possible to select one combination of heads and tails which will be most likely to appear first. All you need to do is drop the third of the bettor's calls and join the opposite of the second call to the front. If your mark called HHT, for example, you would transform it from HHT to HH to THH for your call. The odds are always in your favor! (For precise odds, see the Answers section at the end of this Chapter.) One last stunt involves coinshunting, this one the
31
changing of two rows of coins to a hexagon. Form the two rows in Fig. 24. Now move coin 4 so that it touches 5 and 6, move 5 to touch 1 and 2, and move 1 to touch 5 and 4. Finally, arrange the coins with the top row to the right (Fig. 25), and challenge your opponent to duplicate what you have done in the same number of moves.
Fig. 2·4
Fig. 2·5
12. Lastly, here are some puzzles to confuse and confound: a. You arrange 12 cards, from ace to queen, in a circle. Ask your friend to think of one, but not to tell you which card he has selected. Inform him that you will determine his card by tapping on the cards around the ring. He should count your first tap as 1 above his chosen number, the second 2 above, and so on. When his count reaches 20, he should tell you to stop, and you will both be surprised to find that you are right on top of his cardif you started counting on the right card. What card is this? b. Rearrange 10 of those 12 cards face down to form a box 3 cards by 4 cards, with the center 2 spaces empty. The problem here is to select any facedown card, jump 2 cards in a clockwise manner, and turn over the fourth. Continue doing so until 9 cards are face up. On which card should you start, and what method should you use? c. Draw the star in Fig. 26 in one line and fill in the numbers. The puzzle here is to place a coin on any intersection, jump one intersection, and cement the coin to 32
EUREKA!
the third intersection, always starting and ending on empty spaces, until nine spaces are covered. How can you do this so that a move is always possible and so that the puzzle can be solved? Happy shunting! 2
1 ~+r"7 3
4
5 Fig. 26
Answers
1. If your numbers are N and X: 4N+X 1     X
2
2
X X
2N+2 2
2N
2
2
==N
2
2. If X is your friend's number and y yours, the algebraic expression is:
X  [99  y) + X  100 + 1 ] X  (99  y + X  99) X  (X  y)
=
y
Note that the sum of two numbers less than 100 cannot be more than 198, and that the restrictions on the picking of 33
the numbers means that the sum must be more than 100, so the first digit of the sum of the friend must be 1. Others along this line: You can also ask a friend to throw two dice until he has thrown two different numbers. Then beg him to double the number on one of the dice, add 5, multiply by 5, and add the number of spots on the other die to the product. If you are now told the result, you can determine the two numbers by subtracting 25 and separating the result into two digits. Again mathematics triumphs! Another stunt concerns casting out 9s, or so it seems. Ask a friend to write down a threedigit number and subtract from it the sum of the digits. Tell him to cross out any digit and tell you the sum of those remaining. Merely by subtracting from the next highest mUltiple of 9, you can tell him the digit he crossed out. These are rather easy to devise, and you can no doubt create grander illusions yourself. 3. A number such as 141,414 is divisible by 10,101. As 10,101 is divisible by 7 and 13, 141,414 is, too. A number such as 1414 is divisible by 101; as 101 is prime, this number must be divisible only by 101 and 1. 4. Merely note the fourth number from the bottom, here 50. Multiplying this by 11 gives the sum. Or multiply the first number by 55 and the second by 88 and add the products. 5. The sum of the top faces is 11. The sum of the top face and bottom face of any die is 7, and 7 + 11 = 18. 6. Just subtract the top face from 21, so the answer is 16. For N dice, the sum is 7N  (dots on top face). 34
EUREKA!
7. In the example, 3542/11 = 322; 322  7 = 315; 31519 = 35. Generally, for faces A and B: ([1000A + 100B + 10(7  A) + (7  B) ]/11}  7
=:...·'9'''"~
=
lOA + B
the original number. S. No! He could take your $2, say, "I lose," and hand you his $1. You win the bet, but lose a dollar.
9. I certainly hope not. The buyer will have purchased his own dollar bill for 50 cents. 10. All you need to do to acquire untold sums of money is put down an odd number. Adding an odd number changes an odd to an even, and an even to an odd. Similarly, if you want to lose, you should put down an even number. 11. The odds to the cointossing are as follows: HHH to THH, 7 to HHT to THH, 3 to HTH to HHT, 2 to HIT to HHT, 2 to THH to 'ITH, 2 to THT to 'ITH, 2 to 'ITH to H'IT, 3 to TIT to H'IT, 7 to
1 1 1 1 1 1 1 1
12a. You should start on the 7 and count backward. b. You can start anywhere, but each new start must be three cards counterclockwise to the previous one. c. The secret is to cement each coin to the spot where the previous one started. For instance, you might move from 3 to 10. The next coin must start at 7 and end at 3. 35
3.
A Score of Games _ _ _ _ _ _ _ __
The field of mathematical games is an especially rewarding area of recreational mathematics. Even old, outworn games can be of interest when introduced in new forms.
1. Take, for instance, tictactoe, a game in which two opponents place markers (classically Xs and as) on a 3 X 3 grid in an attempt to get three in a row. After an hour's analysis and play, anybody can become an unbeatable master or force any game into a draw, just by learning two or three simple rules (see Answers). Just one interesting variation is toetactic, in which the idea is to force the other player to get three in a row. How do you win here? Tictactoe on a larger scale is also intriguing. Gomoku, played in the Orient thousands of years ago, is just such a game. The idea is to place five counters in a line on a 19 X 19 board. Players take turns placing from an unlimited supply of counters. Positions which force a win are lines of four open at each end, and two open lines of three. Though experts are of the opinion that the first player can force a win, the game is nevertheless challenging enough to be popular in all parts of the world. Another way of increasing the size of the game is to add more boardsone on top of the otherto form a cube. Here, winning is getting a row on any layer or along any of the lines. The number of possible winning lines, in fact, is given by the formula: (k + 2)"  k"
2 36
EUREKA!
where k is the number of cells on each side and n the number of dimensions. Curiously, the first player can't help winning on the 1 X 1 X 1,2 X 2 X 2, or 3 X 3 X 3 boards. On the 4 X 4 X 4 board, playing is a little more difficult. If you are interested in investigating the strategy further, many manufactured versions are marketed; one model for the crafty is illustrated in Fig. 31.
Fig. 31
DID YOU KNOW THAT ... 25 X 9 2
=
2592?
2. Another exciting variation of tictactoe is played with moving counters, characteristically three to each player. Turns alternate for placing and, if neither player 37
has yet won, moving the counters commences. Usually, only moves along the orthogonals or perpendiculars are permitted, but variations which allow moves along the main diagonals (corner to corner) and/or the short diagonals are known. In a French version, a counter may be moved to any empty cell. Assuming moves along the short diagonals are not allowed, how can the first player force a win? DID YOU KNOW THAT ... The average of 4 and 5 is 4.5? This is unique! Safragat, originating in the Sudan, is another favorite for rainy days. Here four counters are used by each player, and the object of the game is different. Placing is as normal, but the play brings a catchif a player moves so that she traps her opponent's piece between two of her own, that piece is removed from the board. (However, moving a piece between two belonging to the opponent is perfectly safe.) A player loses if she can't move or if she has taken fewer pieces from the board at the end of the game, when no more captures are possible. Try playing on a 5 X 5 board (the game here is known as Sipu) or even a 6 X 6 board with two unoccupied spaces. Teeko, invented by John Scarne, is a deceptively simple game that's easier to play when watching. Players take turns placing and then moving four counters on a 5 X 5 board. Winning positions are four in any line, or four in a box, as shown in Fig. 32.
38
EUREKA!
.~ 0
0 4~
""' 
   .Fig. 3·2
Altogether there are 44 different winning positions, or 58 if any size box is allowed. I strongly recommend Teeko. Nim is another favorite mathematical game that has appealed to many in a wide variety of forms. All, however, concern two players who take objects from an arrangement of piles according to certain rules with the intent of forcing the opponent to take the last object, although sometimes this last part is played in reverse. The simplest form of nim (supposedly from an old English verb meaning "to take away") is played with a single pile of counters. Here, players alternate and take from 1 to m counters, where m is previously determined. For those who wish to win, make sure your opponent must take from a pile of n(m + 1) + 1 counters, where n is any number (and if there are no objects handy, use digits on paper; try to keep from landing on or over 100, for example). This formula was successfully used by a child swindler in John D. Fitzgerald's novel, The Great Brain Reforms; the swindler was a brilliant youth who claimed to hypnotize tin cans so that his opponent would always pick the last one. 39
3. In its more advanced state, nim is played with several piles of counters, classically 3, 4, and 5, or 3, 5, and 7. Players can take up to all of the coins in any single row, the purpose being to force the opponent to remove the last counter from the board. A startling discovery, made around the turn of the century, showed that nim could be generalized into a game with any number of piles and any number of counters per row. A simple strategy enables anyone to playa perfect game. To determine if a position is safe or unsafe (no chance of immediately losing or a chance of losing), simply add up the binary values of each row. Binary, which is base two, is nothing more than the writing of numbers using powers of 2. In the 3, 4, 5 game, the 3 is equal to one 2 and one 1, so its base two representation is 11; 4 is equal to one 4, zero 2s, and zero 1s, so its binary representation is 100; 5 is then equal to 101 base two. Thusly, the sum of the binary representations of 3, 4, and 5 (equal to 11, 100, and 101) is 212. As this sum contains an odd digit, the position is unsafe. To achieve safeness, all you need to do is get rid of the 1 in the center column, making all the digits in the binary sum even. This can be achieved by taking 2 counters from the 3pile. That would leave 1 in the 3pile, and 1 + 100 + 101 = 202, a safe position. Is starting with rows of 2, 3,4, and 5 safe? How would you make a starting position of 4, 5, 6 safe? For playing purposes, you can use your fingers for the powers of 2 up to 1024. Raise a finger for an odd number (that is, a 1) and lower it when it becomes even. The position can be determined to be safe or unsafe by adding each number over the others this way, and determining what you need to do to lower all of your fingers. 40
EUREKA!
In one fascinating variant of nim, players can take from any number of rows up to k. Surprisingly, the same binary analysis still holds; a safe position is one in which every column of numbers totals a digit evenly divisible by (k + 1). Nim follows this rule; players take from one row, so the binary representations must be divisible by 2 to be safe. A game in its own right, Even and Odd also concerns taking counters from a pile, but the object is to end with an odd amount of counters at the end of the game. If m is the maximum number of counters allowed and n any integer, then the player with an odd amount of counters loses if he must draw from 2n(m + 1) + 1 or 2n(m + 1) + 1 + m counters. A player with an even amount of counters likewise loses if he must draw from 2n(m + 1) or 2n(m + 1) + 2 + m counters. Another game similar to nim is the Australian game of 31, played with 24 cards, 4 each of values 1 through 6. The idea is to turn over cards alternately, winning if the score is 31 and losing if it is more than that. The key is to land on the numbers 3, 10, 17, 24, and 31but there is a catch. Consider the following sequence: You open with a 3, and your opponent follows up with a 3. You turn a 4, reaching 10, and he takes a 4 also. You then take a 3, and your opponent follows with a 3, followed by another set of 4s, bringing the total to 28. When you victoriously reach for a 3 to bring the total to 31, you will find there are none to be found. Beware of this! 4. Another variant of nim is Daisy, played with a ring of 13 petals, the players plucking 1 or 2 adjacent petals at a time, the winner plucking the last. How can the second player always win?
41
The Tac Tix board, from a game invented by Piet Hein, is shown in Fig. 33. Like multirow nim, a player can take up to 4 adjacent counters in any row in any direction, the loser taking the last counter. It is unknown which player can force a win, though the first player can always win on oddordered boards (ones with an odd number of counters on a side) by taking the center counter. You might like to try playing on a 6 X 6 board. I

I
000
0
o
0
0
0
0
0
0
0
o
0
0
0

Fig. 3.3
QUICKIE
5. Two sisters decide to split a cake, 20 inches in circumference, between them. Jane suggests that they cut slices of 1 or 2 inches in turn, and Cynthia replies with the proposal that the last to cut should wash the dishes. If Cynthia cuts first, what size piece should she take to avoid washing? Another realm of mathematical games is connecting games. One rather prominent one, also invented by Piet Hein, is called Hex. 42
EUREKA!
Fig. 34
Fig. 35
Played on diamondshaped boards like the two in Figs. 34 and 35, the idea of Hex is to connect adjacent hexagons (or linked dots on a board of triangles) so that an unbroken 43
line extends from one side of the board to the opposite side. The corners can be used by either player. To learn the strategies of Hex, play games on boards of increasing size. The first player can win on the 2 X 2 board, on the 3 X 3 by playing in the center, and on the 4 X 4 by playing in the four middle spaces (if the move is anywhere else, the second player can force a win). This advantage of the loworder boards carries over to the 11 X 11; the first player has a slight advantage, but the edge is very easy to lose. One misplaced play can give the game to the second player. Of course, it does work in reverse, and can be even more disastrous for the second player. Try it and see. David Gale has devised a delightful dotconnecting game called BridgIt, or Gale. It is played on a board which consists of two interlocking grids, each 5 X 6 dots (though you may prefer to play on a larger board), as shown in Fig. 36. One player joins the boxes, and the other joins the dots. The idea is to connect one side of the grid to the other along the long axis without crossing lines. This can never end in a draw. Though the first player on any size board has the advantage, the game is nevertheless challenging and recommended. The game of Boxes is probably familiar to you. A set of 16 or 36 dots is set out in a square. The players alternate in connecting any two adjacent dots. A box is completed if a player draws the fourth side; that player gets credit for the box and takes another turn. The winner is the player with the majority of boxes at the end, when all the boxes are accounted for. Boxes lends itself easily to gambits, one player giving up a box or line of boxes in exchange for more later. One particularly interesting variant of Boxes is Triangles, played on a board the shape of an equilateral triangle. The 44
EUREKA!
•
o
o o
o o
•
•
o D • • o o • • o D • • o D • • o o •
•
o • • o o o • • o o o • • o o o • • o o o D
D • • • •
•
•
•
•
•
Fig. 3·6
purpose is to amass triangles. You might want to compare the skill needed for each. One other classic math game is Mastermind, simple in principle but hard in practice. One player arranges four of six possible colored pegs (or letters or numbers in place of the pegs) in a line behind a small hood. The second player, the "code·breaker," arranges another four of the six on a different part of the board in an attempt to determine the arrangement of the hidden four, the "code." The first player then compares the guess to the actual code. A cor· rect color in the correct row receives a black peg; a correct color in the wrong row receives a white peg. For instance, the second player could receive two blacks and one white, meaning that he had three colors correct, but only two were also in the right row. A set of four blacks means that the second player has deduced the code.
45
Mastermind can be played with a code containing more than one of each color, with four of eight colors or six of nine, with digits, or with entire words. Try to invent your own games. Use such common objects as cards or dominoes, or play your games on undrawable hypercubes. The selection here is just a small sampling.
Answers
1. Tictactoe: There are only three opening moves, as shown in Figs. 37 through 39. The dotted Os are the necessary responses to each X move. Any of the four positions in Figs. 38 and 39, followed by rational playing, will lead to a draw. r\
r"\ ... J
r"\ ... .1
Fig. 37
r"\
X
r"\
\...1
X
r\
\..1
... .1
Fig. 38
\..1 r"\
r"\
\...1
\..1
X
r"\
\..1
Fig. 39
Toetactic: The second player has a decided advantage. The first player can force a draw only by first playing in the center and then playing opposite the second player. 2. The first player, to win, should place in the center. If you number the spaces as shown in Fig. 310, two lines of play follow:
46
EUREKA!
1 2 3 4 5 6 7 8 9 Fig. 3·10
a.X 5 4 9 4 to 7 5 to 8
0
b. 5 1 3 1 to 4 4 to 7
6 9 2 any move
3 6 1 any move
Note that both hold even if moves along the major diag· onals are permitted. 3. The nim position 2, 3, 4, 5 is already safe. The position 4, 5, 6 has binary values of 100, 101, and 110, adding to 313. A safe move is removing 1 from the 4·pile, 3 from the 5·pile, or 5 from the 6·pile. 4. The second player should always keep the position symmetrical, taking petals directly opposite the first. 5. Cynthia should cut a l·inch piece. Her next cut should bring the total of cut cake to 4 inches, and then 7, 10, 13, 16, and 19 inches. Jane must then wash the dishes, but she's probably had 13 inches of cake.
47
4.
The Magic's There _ _ _ _ _ _ __
Magic squares to me are more magic than square. It seems incredible that the numbers from 1 to 16, say, can be arranged in a 4 X 4 square so that the sum of the numbers along any horizontal line, vertical line, or long diagonal is a constant, or that the numbers from 1 to 100 can also be so arranged. The simplest magic square is the 3 X 3, termed the order3 square (Fig. 41). In China, this square is called the loshu. It can be found on amulets, perhaps because the loshu is the only order3 square possible. The seven others that can be formed are either rotations or reflections of it.
8 3 4 1 5 9 6 7 2 Fig. 41
There are variations on the square, however. An order3 square composed wholly of odd numbers can be formed by substituting the 5, say, with the fifth odd number (Fig. 42). The even square (Fig. 43) can be formed the same way. 48
EUREKA!
15 5 7 1 9 17 11 13 3
14 4 6 0 8 16 10 12 2
Fig. 4·2
Fig. 4·3
There is only one order·3 square, but there are two anti· magic squares. In these oddities (shown in Fig. 4.4), no two sums are the same. These are rookwise antimagic squares; that is, you can connect the numbers by moving like a chess rook, either vertically or horizontally.
9 8 7 2 1 6 3 4 5 Fig. 4.4
QUICKIE Warren Peace was thumbing through a copy of Eureka! one day, when he encountered a rather intriguing problem. The diagram showed a 3 X 3 box with a lone 8 in it (Fig. 4·5). "Place a different number in each of the remaining cells so that each line of 3 adds up to 15," said the book.
49
Warren did it. Can you? See the answer at the end of this Chapter.
8
Fig. 4·5
There are 880 order4 squares. The simplest one can be constructed by overlapping two squares (Fig. 46), but it's much easier to write the numbers 1 to 16 in order in a square and switch the 1 with 16,6 with 11,4 with 13, and 7 with 10. Voila!
4 2 1 3 1 3 4 2
3 2 2 3
1
4 4 1
+
12 4 8 0
0 8 4 12
0 8 4 12
12 4 8 0
16 2 3 5 11 10 9 7 6 4 14 15
13 8
12 1
Fig. 46
Interestingly, the square is still magic if the middle two columns are switched. What's produced is shown in Fig. 47. This square appears in Albrecht Durer's engraving, Melancholie, and the date of the engraving, 1514, appears in the bottom row. The magic constant appears here more times than you can shake a stick at: All lines add to 34; the four corners sum to 34; the five 2 X 2 squares, in the 50
EUREKA!
corners and in the center, all add up to 34; the line 3 + 5 and its opposite, 12 + 14, sum to 34, as do 2 + 8 and 9 + 15, and the sums of the squares of the pairs also match.
16 3 2 13 5 10 11 8 9 6 7 12 4 15 14 1 Fig. 4·7
A small rearrangement of the cells yields the square in Fig. 48. This square has all the properties of DOrer's square. In addition, any box of four sums to 34, and a "broken diagonal," such as 3, 5, 14, 12, also sums to 34.
15 10 3 4 5 16 14 11 2 1 8 13
6 9 7
12
Fig. 4·8
It is this last property of the square that makes it a diabolic (or panmagic, or pandiagonal, or Nasik) square. Almost half of all order4 squares are diabolic. These
51
squares are the very devilthey don't change no matter what is done to them. They remain diabolic if they are rotated, reflected, rearranged so that the top row becomes the bottom, rearranged so that the lefthand column becomes the righthand, or rearranged according to a mystic pattern handed down from mentor to pupil through the years that involves switching the cells in cycles of three. So far the magic constant has been determined by trial and error. There is a formula, though, that gives the constant and thus helps in forming squares: (n 3 + n)/2.1f n is 3, the expression gives 15, for example. The constant for order5 squares if 65; knowing that fact is important, since there are more than 13 million order5 squares. Of these, about 28,800 are diabolic. They can be formed by permuting four nonequivalent squares. The square in Fig. 49 is symmetrical as well as diabolic. Every pair of squares opposite around the center adds up to 26, twice the center. The loshu, by the way, is also symmetrical.
18 1 14 22 24 7 20 3 5 13 4 9 6 19 2 15 4 12 25 8 16
10 11 17 23
Fig. 49
52
EUREKA!
Unfortunately, diabolic squares are not possible in evenorder squares not divisible by 4, though normal squares, of order6, say, can be constructed. Order8 squares can be constructed, too. Figure 410 is an odd one.
123 121 119 117
13
15
19 107 105 103 101
29
31
39 41
43 83
81
79 77
53 55 57
59 67
65
63 61
69
51
49
47
45
85 87
97
99 27
1
17
3
95 93 37
113 115
11
71
73
89 91
35 33
21
109 111
5
125 127
25 23 9
75
7
Fig. 4·10
Ben Franklin constructed the square in Fig. 411, in which each row adds up to 260, and each halfrow adds up to 130. The sum of any box of four is 130, as is the sum of
53
any four numbers equidistant from the center. Following any full dotted diagonal up and down also gives 260.
Fig. 4·11
The 8 X 8 square resembles a chessboard, so somewhere along the line someone asked himself if a knight, which jumps two squares up or down or to the side and one perpendicularly, could cover each cell of a magic square in order. Leonhard Euler created such a magic square with a constant of 260; stopping halfway on each horizontal or vertical line gives 130. Henry Dudeney created a magnificent square. Not only can the knight jump from the 64 to the 1 to restart its tour, but the diagonals are also nearly correct256 and 264.
54
EUREKA!
These large squares are not as hard to create as you might think. There is one very interesting method that produces a concentric, or bordered, magic square. These squares stay magic as their outside rows are repeatedly dissolved until the core of an order3 or order4 square is reached. The symmetrical order5 square in Fig. 412 is bordered, as is the order6 square in Fig. 413.
19 2 4 16 18 11 21 12 3 24
20 9 13 17
1 23
14 22 15 8 10 5 6 25 7
Fig. 412
36 2 3 7 32 29 26 13 12 23 27 15 20 21 18 9 19 16 17 22 4 14 25 24 11 6 35 34 30 5
31 8 10 28 33 1
Fig. 413
55
So far these squares have been composed wholly of consecutive integers, odd integers, or even integers. Is it possible to create a magic square composed wholly of primes or consecutive composites? Yes, it is, and these squares truly are incredible. Dudeney tackled the composite square. He found that in the interval 114 to 126, all the numbers are composite, so he created the square in Fig. 414. His prime square, oddly enough, sums to 111, the constant for order6 squares (Fig. 415). Bergholt and Shuldham topped him with the simple square in Fig. 416. Prime squares have been found past order12. Once in that range, though, the magic constants become huge.
121114119 1161181
67 1 43 13 37 61 31 73 7
Fig. 4·14
Fig. 415
3 71 53 11 17 13 29 7
5 37 41 19
Fig. 4·16
56
EUREKA!
23 1
31 47
An odd variant of the magic square is the magic cube, consisting of n magic squares of ordern piled so that the sum along any horizontal line, vertical line, or long diagonal is the same. The simplest is the order3 cube (Fig. 417), but cubes of order6 have been constructed. Fans of Rubik's Cube may (but I doubt will) enjoy placing these numbers on the cube in place of the colors and solving it by solving the magic cube.
10 24 8 23 7 13 9 11 22
26 1 15 3 14 25 13 27 2
6 17 19
16 21 5 20 4 18
Fig. 417
Other interesting variations on the magic square are the multiplication, division, addition, and subtraction squares. In the multiplication square in Fig. 418, every line multiplies to 216. If the diagonal corners are switched, it becomes a division square, in which the product of the end cells divided by the middle always results in 6. The normal magic square can be changed into a subtraction square by switching the corners; the sum of the edges minus the middle cell is always 5.
12 1 18 9 6 4 2 36 3 Fig. 418
57
The square can also be used to force certain numbers. Let's say you wanted someone to select, perhaps for some part of a larger trick, the number 43. The first thing to do is to decide what order square to use, and then split the 43 into twice that number of parts. An order4 square can be used, so 43 is split into 1, 3, 4, 5, 6, 7, 8, 9. The numbers are arranged randomly on the outside of the square, and their sums on the inside (Fig. 419).
+ 6 4
1 8 5 9
7 14 11 15
5 12 9 13
3 7 4 8 11 15 8 12 12 16
Fig. 419
Any number in the 4 X 4 square is circled, and all the others in its horizontal and vertical rows are crossed out. When four have been chosen, their sum is 43. This works because each of the original parts is included in one of the chosen numbers, so the sum has to be 43. Magic squares are used more in this hardertofigureout trick. Arrange the first nine cards of a suit in the order ace, 8,2, 7, 3,4,5,6,9, so that the ace is on top of the packet. Have a spectator riffle these cards through the deck, and then pull them out again. This should not disturb the order. Now for a game of tictactoe. Split the packet, faces up, 58
EUREKA!
so that your left hand holds the lower six cards and your right the higher three. Place the 5 in the center of the board and ask your opponent for his move. If it is a corner square, put the righthand cards on top of the lefthand cards; otherwise, put them beneath. Now put the packet face down on the table and ask the spectator to place the top card in the square he picked. If it is a side square, ask him to place your card in the adjacent corner. If he places his first card in the corner, place it on the opposite side square. The rest of the moves are forced. When all the cards are overturned, a magic square is revealed! This is just one of the many interesting things about and uses of magic squares.
Answer
Figure 420 shows Warren's solution. Oddly enough, 8 is the number of lines on which the constant 15 can be found.
4~ 8 2~
3 5 7 7~ 2 5~ Fig. 420
59
5.
Rubiquity _ _ _ _ _ _ _ _ _ _ __
The Rubik's Cube is an innocentlooking, often maddening puzzle that has caused millions of wouldbe solvers to go quietly crazy. Sweeping the world is the rather pleasurable disease of cubitis, a mysterious and highly contagious malady that is both caused and cured by contact with this guaranteed hairtearer, the separate brain children of both Hungarian architecture teacher Erno Rubik and Japanese engineer Terutoshi Ishige. Each side of the 3 X 3 X 3 cube (Fig. 51)which is itself composed of 26 smaller cubies grouped around a jacklike spindleis originally a different color, but each face can be twisted or twiddled so as to produce wild, multicolored messes.
Fig. 5·1
The basic problem is to rearrange the cubies so that each face is a solid color again. Those who haven't been stumped 60
EUREKA!
by Rubik's Cube can get a real appreciation of the skill involved in solving one with the news that there are more than 43 quintillion (or 43 X 10 18 ) possible arrangements of the cubiesand only one is the solved position. Here's how that discouragingly large number is achieved: There are eight corner pieces. Each can be twisted three ways. The first factor, then, is 38 . However, because one good turn deserves another, and every negative twist has a matching positive, when seven corners are oriented, the eighth is fixed, so a factor of 3 is removed. The corners can also be arranged in the eight cubicles in 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 ways for another factor of 8!. (The ! is a factorial sign. X! indicates the product of all positive numbers less than X + 1. For example, 5! = 5 X 4 X 3 X 2 X 1 =120, and 8! = 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 40,320.) The edges can also be permuted in 12! different ways, but when eleven are placed, the twelfth is fixed, so a 2 is removed. Also, each cubie can be oriented in two different ways, so a factor of 212 is produced. But it, too, must be reduced by 2 for that last fixed cubie. The product of all these numbers is 43,252,003,274,489,856,000. If you were to look at one of these positions every second without rest, it would take you more than 1 trillion years, or more than 100 times the life of the universe! Easy puzzle, eh? Well, it can be. The key to solving it, usually within minutes instead of millennia, is to separate the solving process into stages, in which the numbers involved are much smaller. But I'm getting ahead of myself. The first thing to do is to get a mathematical appreciation of the cube and a basis for solving it. Part of the challenge in solving the cube is to overcome the difficulty in analyzing it that results because so many things can happen so quickly; it just so happens
61
that a branch of mathematics deals with the cube and what can happen to the cubiesgroup theory. A series of moves on the cube can basically be seen as a group. A system of notation originated by David Singmaster in Notes on Rubik's Cube helps describe the moves. If the cube is placed flat on a desk, the upper face, whatever its colors may be, is called V. The righthand face is R, the left is L, the front F, the back B, and the downward side is D. A positive move is a clockwise move of that face; R moves the righthand face away from the solver, and L moves the left toward the solver. A negative move is indicated by a small slash: V' moves the top face counterclockwise. A move of the back, upper, and right faces, all clockwise, can be represented as BVR. Two moves of the downward face are expressed as D2 , which is equal to (D')2. Notice the power of conjugates: B' undoes B, and FD'R' undoes RDF'. This notation is very useful in passing ideas. If you have a solved cube (if you don't, then turn the top face 45 degrees, insert a screwdriver under one of the top edges, pop it out, remove the others to get to the spindle, and form the solved cube by placing the cubies back in their appropriate spotswhich is how I solved my first cube), you might care to see what you get with RL'FB'VD'RL' or (R2D 2V 2L 2V 2)D(L2F2B2L2F2B2 )D. The notation can name every cubie. The edge between the V and R faces is called ur, and the corners next to it are urf and urb. This system comes in handy in the solution. Here is the only labelled joke in this entire book: Q: What is Erno Rubik's favorite theatrical play? A: Rubik's favorite play is, of course, Karel Capek's Rossum's Universal Robotsalso called RUR! 62
EUREKA!
The cube can always be solved if it is in one of those 43 quintillion positions. If some joker were to put an edge in backwards, though, one of eleven new universes of 43 quintillion more positions will be opened. Even in the "worst" orbit, however, the cube can be solved except for a twist in one corner and an exchange of two edges, one of which is flipped. Otherwise, the cube can be solved through the use of an orderly system. Certain cubes are selected and placed correctly in this algorithm. What surprises many beginners is the fact that the partially solved cube must be messed up and restored several times along the way. This is a hard lesson to learn. My system, along with many others, uses AHA' moves. Let's say you have a nice operator, a move that will flip two edge cubes if they're in the right cubicles. Perform the series of moves A (which could be B'DU) to get the cubies in those two magic spots, perform your operator H, and perform A' (which could be U'D'B) to get the cubies back. AHA! Those edges are flipped! The algorithm presented here proceeds according to these stages: top center; top corners; top edges; equator centers; bottom corners; equator edges; and bottom edges. Top Center
This is the easiest stage. Just choose a nice color and put its center on top. I usually start off with orange. Top Corners 1. Simply get an orange corner label up next to the center label. This requires at most a halfturn. Turn the entire cube so that the corner is at ufI. (Ha! You didn't believe me!) Two or more oranges should be on top now. 2. Note the color on the f side of ufl. There is only one 63
other corner cubie with labels of orange and this color, and it goes at ufr. Find it. If that cubie is at ufr and the orange is up, go straight to step 5. 3. If the cubie is in the top layer, but not correctly placed, turn the entire cube so that it is at ufr. Perform R'DR and return the cube to its original position. 4. Turn the cubie in question to rfd by moving the D face. If the orange is on the R face, perform R'D'R to get it on top. If it is on F, perform FDF ', and if it is on the bottom, perform R'DRFD 2 F' or D 2 R'D'RFDF' . 5. The next cubie is now correct. If all four orange corners are up and correctly placed (Fig. 52), go on to the next stage. Otherwise, make a Uturn and treat the cubie now at ufl as the one you originally picked . Go back to step 2please.
Fig. 52
Top Edges 1. The object of this stage is to complete the uppermost layer, so that the top face is all orange and the trimmings around the sides are solid blocks of color. If all four top edges are correctly placed, colored, and oriented, go to step 7.
64
EUREKA!
2. Turn the cube so that one edge cubie that is wrong is at uf. Note the colors of the f faces on the corners at fur and ful (they should be the same). Now locate the edge cubie with labels of orange and this color. 3. If it is at uf but not correctly oriented, perform FDU'R'U to get it into the middle layer, and go to step 5. 4. If the cubie is in the top layer but not correctly placed, turn the cube so that it is at uf and perform FDU' R'U and move the cube back. 5. Find that cubie. If it is in the middle layer, turn it to fl. If the orange is at L, perform U'R'UD'F. Otherwise, perform U'RU 2 D 2 L'. Go back to step 1. 6. If the cubie is in the bottom layer, turn it to fd. If the orange is on the bottom, perform RL'F 2 LR'. Otherwise, perform F'DU'R and go back to step 1. 7. Congratulate yourself. You've solved a layermore than onethird of the cube! Equator Centers Turn the middle layer so that the centers match the trimmings of the top layer, as in Fig. 53.
Fig. 53
65
Bottom ComersPosition 1. Turn the cube so that the orange is on the bottom. Choose a corner on top and correctly position it by turning Us several times so that its colors match the equator centers. Turn the entire cube so that the orange is at the left and the cubie is at rdb. 2. The orange should now be at the left, and a correctly placed and possibly correctly oriented corner should be at rdb. 3. Examine the three other corners on the R face. a. If they're all correctly placed, skip the rest of this stage. b. If they need to be switched clockwise in order to make the equator centers match the colors, perform (R'D'R)U(R'DR)U'. c. If they need to be switched counterclockwise perform U(R'D'R)U'(R'DR). This is the conjugate of the above. d. If two need to be switched diagonally, perform D'R'F'RFDR'. If two next to each other need to be switched, select one of them as your new base. Go back to the start of step 3. 4. Don't read this step.
Bottom Comersorientation What started off as the bottom, the red (if red is opposite orange) should now be turned to the top. 1. If all red labels are on top, go on to the next stage. 2. If no reds are up, perform (R'DR)(FDF')U(FD'F') (R'D'R)U'. This should bring one to the top. 3. If only one red is up, turn it to ufI. a. If there is a red at the f of fur, perform (F'LF)R' (F'L'F)U2(R'D'R)U2(R'DR 2). 66
EUREKA!
b. If the red is at the r, perform (R 2 D'R)U 2 (R'DR) U 2 (F'LF)R(F'L'F). c. Go to the next stage. 4. If two reds are up, turn them to ulf and ulb. (If they are diagonally opposite, turn one to ufi. Follow the directions below, but substitute U 2 for U and U'.) a. If the red is at the f of fur, perform the meson, (R'DR)(FDF')U(FD'F')(R'D'R)U'. b. If the red is at the r, perform U(R'DR)(FDF') U'(FD'F')(R'D'R). 5. Rejoice. Equator Edges
This is the hardest stage to explain. The basic move is a threeedge swap between fd, fu, and bu (if the orange is at left or right). The possibilities are too many to write down, but here's the general set of steps:
1. In the equator (actually the Prime Meridian in this case), find an edge cubie that's incorrectly placed, and turn the cube so that it is at ufo 2. Put the cubie that should be at uf at fd. It might help to get it to either db or dr first. Call this move X. Don't hesitate to write it down. 3. Put the cubie that's in the space that the cubie at uf should be in at ub, so that fr can go to fu can go to ub. Call this move Y. 4. Perform RL'U 2 LR'F 2 • 5. Now perform X'Y'. 6. Go back to step 1 until all the equator edges are correctly placed, even if they're not correctly oriented.
67
Bottom Edges
Turn the red so that it's up. Either all four edgecubies have to be switched, or only three have to be. 1. If uf should be at ub and ul should be at ur and vice versa, perform R2L2DR2L 2U 2R2L2DR2L2. 2. If all four are wrong, turn the cube so that ur should be at ufo Perform LRB(L 2U 2 )3B'R'L'. 3. If three are wrong, turn the correct one to ul. a. If the cubie at uf should be at ub, perform (R 2D') (RL'U 2LR'F 2 )(DR 2). b. If the cubie at uf should be at ur, perform (R 2D') 2 (F RL'U 2LR')(DR 2 ). 4. One hundred percent of the cubies are now correctly placed! Edge Orientation
1. Turn the entire cube so that one flipped edge is at ufo 2. Turn faces so that another flipped edge is at ub. 3. Perform either a or b: a. FUD'L2U2D2RU2R'D2U2L2DU'F'U2 b. (LR'F)(LR'D)(LR'B 2 )(RL'D)(RL'F)(RL'U 2) 4. Undo step 2. 5. Now go back to step 1. 6. Tell everyone within shouting distance: You've solved the cube!!! More Challenges
Now go for speed, make pretty patterns, try odd colorings or cubes with unusuallyshaped cubies, solve the new 4 X 4 X 4 cube, or try to solve the cube in the mirror. Good luck!
68
EUREKA!
~oNiftY Numerics 6.
Palindromesemordnilap _ _ _ _ _ __
A palindrome is a word or phrase that reads the same forward as backward, such as rotator or "Madam, I'm Adam" (to which she aptly replied, "Eve"). In mathematics, numbers like 39,593 and 123,454,321 are palindromes. It has been conjectured that the following process would always yield a palindrome for any starting number: Add any number to its reversal. If the sum is not a palindrome, then repeat the process on the result. The number 459, for instance, takes two steps to become palindromic, and 549 takes five. Well, the conjecture is nearly rightalmost 98% correct. The process above was tested on all the positive integers less than 10,000, and 97.5% of them produced palindromes in 24 or fewer steps; however, the remaining num69
bers failed to become palindromic in the first 100 steps. The innocent 196, in fact, fails to produce a palindrome in 37,303 steps (and contains some 15,500 digits at that stage! ). Palindromes seem to pop up frequently. The product of 7, 11, and 13, successive primes, is 1001, a palindrome. Palindromic pairs of primes, such as 13 and 31, or 37 and 73, are not uncommon. Should palindromic primes like 151, its own reversal, be counted? Any palindrome multiplied by 11 (another palindrome) results in a new palindrome. A palindrome can be the cube of a prime only (as in 1331 = 11 3 ). Perhaps you can find some more relationships like this. Here are a few problems involving palindromes:
1. Rhoda and Phil O. Dendron, both parttime math tutors, found that their earnings over a twoweek period were reversed for each week (Rhoda might have earned $58 the first week and $85 the second). If each earned $165 in the fortnight, what were their weekly earnings? 2. V. Hickle, a cautious driver, was out for an afternoon drive when she noticed that the odometer showed 45,954 miles. "I bet it'll be a long time before a palindrome turns up again," she said to herself. Yet, two hours later, the odometer showed a new palindrome. How fast was she traveling during those two hours? Answers
1. One of the duo earned $78 and $87, and the other earned $69 and $96. 2. Victoria Hickle traveled at precisely 55 m.p.h. for those two hours, covering 110 miles. At the end of the trip, the odometer showed 46,064 miles. 70
EUREKA!
7.
A Pole Vaulter _ _ _ _ _ _ _ _ __
Hap Hazard, worldfamous hunter and sportsman, set up a camp from which to go bearhunting. One morning he found bear tracks outside his tent, and, gun at the ready, started to follow them. Stalking his prey, he traveled 1 mile due south, turned, and crept 1 mile due east, at which point he stumbled across the bear and vanquished his sleepy opponent in a vicious handtopaw fight. After taking the required victor's photographs, he dragged the carcass back to camp, a distance of precisely 1 mile due north. 1. What color was the bear? 2. How may points are there on the globe from which such a journey as Hap's can be taken?
Answers
1. To answer first things first, the bear must have been a polar bear and hence must have been white. The North Pole is the only bearinhabited place from which you can walk a mile south, a mile east, and a mile north and find yourself at the point from which you started. 2. If the South Pole had bears, there would be an infinity of answers to the problem, because there are an infinity
71
of points from which such a walk can be taken (the title hints at this). If Hap had started at a distance of about 1 + 1!(27T), or about 1.16, miles from the South Pole and walked 1 mile south, the walk of a mile east would have taken him all the way around the globe, and the mile north would have returned him to his starting point.
72
EUREKA!
8.
A Timely Switch _ _ _ _ _ _ _ __
Problems involving the transfer of liquids are among mathematics' most enjoyable challengesperhaps people like them because the solver can always try the experiments himself!
1. In a state of global emergency, you are unexpectedly called upon to measure out precisely 4 quarts of liquid hydrogen from a huge tank. However, you have been provided with only a 5quart measure and a 3quart measure with which to save the masses. How do you avert the destruction of the world? 2. As part of a psychiatric examination, you are given a bucket containing 24 ounces of fluid and three bottles, holding 5, 11, and 13 ounces. The test is to separate the original 24 ounces into three drinkable, 8ounce volumes. How do you prove that you are sane and thinking? 3. This one is a little different. You need to remove a delicate, glazed bowl from the kiln in precisely 9 minutes. Unfortunately, there are no clocks handybut you do have two accurate hourglasses, a 7minute and a 4minute. How do you measure exactly 9 minutes?
Answers
1. You quickly fill the 3quart measure with hydrogen 73
from the tank and pour it into the 5quart. Now you refill the 3quart and pour from it into the 5quart until the 5quart is full. Now you dump the contents of the 5quart back into the tank, pour the 1 remaining quart in the 3quart measure into the 5quart, and refill the 3quart. At last, by merely pouring the contents of the 3quart into the 5quart, you have saved the world. 2. First, fill the 11 and 5ounce bottles, leaving 8 ounces in the bucket, and pour all the contents of the 5ounce into the 13ounce. Now pour from the l1ounce into the 13ounce until the 13ounce is full and the l1ounce has 3 ounces. Now refill the 5ounce from the 13 and pour the contents into the l1ounce. Voila! You can reason! 3. Start both hourglasses. When the 4minute runs out, flip it over; when the 7 minute runs out, flip it over. When the 4minute runs out again, flip them both over. When the 7minute runs out a second time, 9 minutes will have elapsed.
74
EUREKA!
350 9.
Fallacies and Logic
If This Is Not a Chapter, My Name Is Raymond Smullyan _ _ _ _ _ __
Let's say that one fine morning you awake to find yourself on an obscure South Seas island. By and by, you discover that the natives all speak Paradox, a language you are familiar with, and are all of two contrasting tribes, one consisting wholly of truthtellers and the other of liars. Unfortunately, it is impossible to tell which tribe an inhabitant belongs to by his dress alone. Now, it also happens that all the natives are rather closemouthed, and have the nasty habit of disposing of anyone who asks them more than one question. For this reason, you consider it best to make your way to the local village, 75
where you will undoubtedly be able to secure transport off the isle. Setting off in what seems to be the right direction, you soon come to a fork in the road, beside which a native is lounging. So, before you read on, can you determine what single question to ask this liar or truthteller, so that you can figure out which branch of the road leads to the village? There are actually several solutions, all along the same basic lines. For example, you could point to one of the roads and ask the islander, "If I were to ask you if this road leads to the village, would you say, 'yes'?" An answer of "yes" indicates that this is the correct road, even if the villager is a liar! The liar would respond "no" to a direct question if the path was the right one, but must lie about his response to your question, giving an answer of "yes." A similar question would be, "If I asked a member of the other tribe whether this road leads to the village, would he say, 'yes'?" Here a "no" answer indicates that the road is correct. A better way of saying this is, "Of the two statements 'You are a liar' and 'This road leads to the village,' is one and only one of them true?" What would you make of a "yes" answer? Again a liar would lie about his being a liar and about the number of true statements, so a "yes" does indicate that you're heading in the right direction. Another stratagem is, "Is it true that this is the way to the village if and only if you are a liar?" A "no" indicates that it is indeed the way to the village. Now suppose that you come to a crossroads with a native conveniently near (actually, you suspect that he may be sitting on the signpost, but you don't want to waste your question asking him to move). You are prepared to ask him the same question that set you off correctly on this road, but "you find, to your horror, that you have forgotten the Paradox words for "yes" and "no." 76
EUREKA!
All you can remember is that "plink" means one and "clunk" the other, but you don't know which is which. Can you still determine which road leads to the village? Yes, you can. Just point to one of the roads and ask, "If I asked you whether the road I am pointing to is the road to the village, would you reply, 'plink'?" Regardless of the tribe of the native or the meaning of "plink" and "clunk," a reply of "plink" indicates that the road is the one to the town. If "plink" means "yes" and the road is the right one, a truthteller would reply "plink," or "yes," to your question. A liar would lie about his lying response; thus he would also say "plink." Similar logic applies if "plink" means "no." The truthteller, for example, would say "plink," or that you were incorrect in saying that the road was not the one to the village. And if the road were not the right one, you would hear the opposite reply in all cases, a "clunk." (What you don't want to hear are things that go "clunk" in the night.) So, if you ask with a "plink" and get a "plink" back, you're on the right road. If you were to come across a particularly quiet inhabitant of the island, you could still determine which branch leads to the village with the question, "Which of these roads leads to the village?" Presumably, a liar would point to all the wrong ones. Unfortunately, he might also be lazy and merely point to one road. Thinking that he was a truthteller pointing out the correct road, you would soon be going in the wrong direction. All of these questions are also subject to misinterpretation. "If I were to ask you if this road leads to the village, would you say 'yes'?" could seem the same as "Does this road lead to the village?" to an especially simple liar, who would helpfully give you the wrong answer. On the other hand, you might come across a particularly clever liar who realizes that he's being tricked into giving the right 77
directions. (This illustrates the difference between an honest liar and a malicious, deceitful liar.) Here are a few problems involving truthtellers and liars:
1. You come across three natives engaged in conversation, so you decide you can probably ask more than one question. You ask the first man (A), the quietest of the lot, what his tribe is, but you cannot hear his low reply. The next, person B, tells you that A said he was a truthteller, but C, the third native, tells you that B is a liar. Assuming that you want your next question to be addressed to a truthteller, to whom should you put it? 2. What do you make of the following statement: "If A, B, C, and D each speak the truth once in three times, and A affirms that B denies that C declares that D is a liar, then what is the probability that D was speaking the truth?" 3. Since your skill as a logician is famed throughout the land, you are confronted with the following three statements: A: "Bislying." B: "c is lying." C: "Both A and B are lying." Who is not lying? 4. One of four crooks is telling the truth. Their statements are as follows: A: B: C: D: 78
"One of us is lying!" "No! Two of us are lying!" "Not so! A full three of us are lying!" "False! All of us are telling the truth!"
EUREKA!
Who is telling the truth? This can also be represented this way: A: B: C: D:
One statement is false. Two statements are false. Three statements are false. Four statements are false.
How many statements are false? 5. Another type of paradox is contained in a problem frequently represented in an eggfinding or mantobehung form. A mathematics teacher, known to be extremely fairminded, told her class one day, "I've scheduled an algebra test for sometime next week. However, I am not going to tell you what day the test is on. You will not know when we are having it until you come in that day for class." And she left. Immediately the class logician stood up and quieted the groans. "It is obvious that she can never give the test without us knowing beforehand." He strode up to the blackboard and chalked down the days of the week from Monday to Friday. "Now, we can't have the test on Friday because, if we haven't had it by the end of next week, that would be the only possible day left," he explained, erasing "Friday" with great relish. "So it can't be on Friday. If we haven't had the test by Thursday, then it would have to be on that day because we have already proven that it can't be on Friday; we would then know that it would be Thursday, so we can cross it out, too." In a likewise manner, he erased Wednesday, Tuesday, and Monday. As a result of this brilliant proof, no one studied for the test. Imagine their surprise Wednesday morning when the 79
exam papers were handed out! (This problem is a classic. I hope you can see the flaw in the logic. One debatable solution is to expect the test every day. Then again, you really don't know when it will be given. Thinking about it only makes it more difficult to figure out.) Come the next test, the teacher decided not to be so clever and just let the class know, though in a roundabout way, that it was sure to have a test of some sort. She told the logician, "If you make a true statement, we will have a small test. But if you make a false one, we will have a large quiz." If you were in his place, could you figure out what statement to make so that it would be impossible for the teacher to give you a test? This leads us into the subject of paradoxes in simple, innocentlooking sentences. Remember the island of truthtellers and liars? Why was it impossible for any islander to say, "I am a liar"? (Not unless he is a normal human, a third type that either lies or tells the truth indiscriminately. Then he could be telling the truth about his lying.) If true, then it's false, which makes it true, which makes it false .... Another sentence like this is "This sentence is false" which is false if true and true if false. This can be changed into "This sentence can never be proved." Try to! P.E.B. Jourdain devised a card which reads on one side, "The sentence on the other side of this card is true," and on the other side, "The sentence on the other side of this card is false." You might like to make such a card and give it to your worst enemy. (That reminds me: Is your worst enemy your best friend? What is your best enemy?) 6. "The number of words in this sentence is nine." Try to construct a sentence equally true but nevertheless saying the exact opposite. It's simpler than you think. 80
EUREKA!
I once made a bet with the captain of the track team, who used to be a good friend of mine, that I could beat him in a halfmile race (his best event, of course). When the day of the race turned out to be the hottest day that summer, I kindly offered to postpone the race, but he wanted to see me suffer. "Well, I really don't quite feel like running today," I said, squirming a little and trying to assume a pleading expression. "Perhaps if you gave me a foot's head start I might." This he readily agreed to. He even laughed at me! "Well, even that's not enough," I said, suddenly pausing. "Hey! I have an idea! Let's just figure out who would win." Again he agreed, assuming a victorious air, so I created an argument that went something like this: "Let's just say that you run twice as fast as I do (which he almost did). When we start the race, I'll be a foot ahead. Therefore, by the time you cover that foot, I'll be half a foot ahead. Right? By the time you cover those six inches, I'll still be three inches ahead. My lead will, of course, diminish to 1.5 inches, then to 0.75 inches, and so on." (And we would never get more than two feet into the race.) Unfortunately, we never finished the argument or started the race, but you do get the general idea of the paradox, first proposed by Zeno, a Greek philosopher, some 2000 years ago. In his version, a tortoise beat the mighty Achilles in a foot (and flipper) race. Another paradox along this line is one most students are familiar with, and I suggest you try it now. Place your pencil down anywhere in the approximate center of this page and draw a light line toward the edge, so that your pencil tip is now precisely twice as close to the margin as it was before. Repeat this until you have reached the edge of the book (even use a ruler if necessary). The fallacy here is mathematical. Each time you move
81
your pencil, the distance covered is half the distance covered the move before. This can be expressed as the . 1 + "8 1 + 16 1 + 32 1 ••• ,t h e sum 0 f wh·IC h approac hes, series "21 + "4 but never quite makes it to, 1. If you are precise enough, you can continue until the end of the universe. (Perhaps it is easier to see that with each move of the pencil, you cover half of the remaining distance; thus you can never reach the full distance, no matter how many moves you might make.) 7. Computational fallacies abound. Consider this one:
9  24 = 25  40 9  24 + 16 = 25  40 + 16 (3  4)2
=
(5  4)2
34=54 1
=
1
Gotcha!
8. What's wrong with this one?
V5=X = 1 +yx 5 x
=
1 + (2yx) + x
4 2x=2yx 2x=yx Squaring:
4  4x + x 2
= X
0= x 2

o = (x x =4 82
EUREKA!
5x + 4
4)(x  1) or
x =1
If we substitute these roots back into the equation, the answer of 1 checks, but 4 gives an odd result:
v"5=4= 1 + V4 1=1+2 (Hint: The above two use the same gimmick.) Here's a neat fallacy involving quadratic equations (don't solve them like this!): (x+3)(2 x)=4
x + 3=4
or
x =1
or
2 x =4
x =2
Correct! In fact, you can make up your own fallacy of this type with your own roots p and q (ignore the x until the final form). Just represent it like this: (l + q  x)(l  p + x) = 1  p + q and solve it as above. Here's another curious generalization:
a2

b 2 = (a + b )(a  b)
so: a + b = a2
b2 a b 
Cancel the a and b, and then the minus signs: a+b
=
a~
f
b~
11th a+b=a+b
Correct!
83
Another set of mathematical fallacies is of the outworn sort:
x=O x(x  1) = 0 x  1= 0
x=l 1=0
where dividing by zero produces the fallacy. A betterdisguised one:
a=b+e a2
ab
=
ab  ae  b 2
ab  ae
=
ab  b 2
a(a  b  e)
=
b(a  b  e)
a2




be
(multiplying by a  b)
be
a=b Another branch of fallacies involves i, the imaginary square root of 1: (y=T)(y=T) =y(l)(l) y=T2 =y'T
1
=
1
The method is just fine except for one fact: (1)(1) = 1, but i 2 = 1; the mistake is thus introduced in the first line.
84
EUREKA!
Also:
yCI=yCI
VV1=vm VI lyCI = yCI/VI (VI)(VI) = (v=r)(yCI) 1 =1 That one's a little harder to figure out! 9. Geometric fallacies are also quite interesting. A neuropsychiatrist devised a mindturning demonstration of a paper triangle whose back has a smaller area than its front. If you dissect a 64unit square as in Fig. 91, and rearrange the pieces as in Fig. 92, a rectangle of 65 square units results! What is wrong? 5
3
/
.......
J
/ 1/ .......
r. ........
3
I' ......
Fig. 91
8 .......
8
5
I'I' ......
5
...... ........
3
3 r....... ........
........
5
I'...
Fig. 92
8
85
10. Here's a small puzzle in logic: Three business partners went out to dinner one evening and ran up a check of $30. Each gave the waiter $10 (and also charged the full $30 to the expense account of each, I might add). Before their change was returned, the manager decided not to charge them for their drinks and returned $5 to the waiter to give to the partners. The waiter didn't think they had tipped him enough, so he gave each of the partners $1 and pocketed $2 himself. Now, each of the three had paid $9, making $27 in all, and the waiter had $2, making a total of $29. What became of the last dollar? Answers
1. The first man (A) must have said that he was a truthteller, because a liar would lie about his tribe. Therefore, B must have told you the truth, and C lied, so your next question should be to B. 2. This question raised a lot of controversy when it was first proposed half a century ago. If a table is formed with all the responses of the four, it turns out that 40 of the 81 possibilities are internally inconsistent. Of the remaining 41, 13 begin with D telling the truth, so the probability of his telling the truth is 13/41. 3. B is telling the truth. If he were lying, then both C and A would be telling the truth. But as C is saying that A is a liar, B must be telling the truth. 4. In the first case, D obviously cannot be telling the truth. Therefore, D is lying. This means that A is true, and then Band C are telling falsehoods. But this would make three false statements, and three liars, which con86
EUREKA!
tradicts C, so A is also lying. Now B could be telling the truth and C lying, but that again contradicts C, so A, B, and D are lying and C is telling the truth. In the second case, D contradicts itself, so it must be false. Logic then proceeds as above. 5. You should say, "We will have a large quiz." If this is true, the teacher will give you a small testbut then you have uttered a false statement, so you should receive a large quiz. But then your statement is true .... 6. "The number of words in this sentence is not nine." 7. The error is in the change from the third step to the fourth. If we substitute values in: 15 = 15 1=1 (_1)2
=
(1)2
1 = 1 This is an example of false square roots. The square root of a number squared is not always that number. For instance, the square root of the square of 1 is 1. 8. Again, squaring or taking the square root of a number can introduce false roots. For instance, the roots of x 2 = 1 are both 1 and 1. Taking the square root adds an answer. In the fallacy, squaring added a false root which must be checked in the original expression. 9. The lengths of the line segments are in the Fibonacci series, where each term is the sum of the previous two terms. The ratio of successive terms approaches cp, a neverending decimal equal to (1 + yI5)/2 or 1.61803398 ...
87
from either side. This means that the rectangle alternately loses and gains area along the diagonal, explaining this paradox. If the square is not to lose any area, the lengths must be cut according to the series 1, cp, cp2, cp3 , cp4, ..•. 10. If you read the problem carefully, you can see that the other dollar went nowhere. Of the original $30, the manager kept $25, the waiter kept $2, and $3 was returned to the patrons (or, if you prefer, the partners paid $9 each for a total of $27, of which $25 went to the manager and $2 went to the waiter).
88
EUREKA!
10.
Thrice Befuddled _ _ _ _ _ __
1. The Pythagorean tripletsKate, Kathy, and Katherineclaimed to be perfect logicians who could instantly deduce all the consequences of a given set of premises. One day Archimedes Anderson, head of his school math team and logician extraordinaire, decided to put them to a test. On the head of each of the sisters he put a hat, red or green, so that no girl could see the color of her own hat. Grinning at his own cleverness, Archimedes asked, "Will each of you please look at the hats of the other two, and if you see a red hat, please raise your hand." All three immediately raised their respective hands. Practically rolling on the ground with mirth, Archimedes said, "If any of you is certain what color hat you have on, please raise your hand." After a few seconds, Katherine raised her hand and correctly named the color of her hat. What was her reasoning? 2. Well, Archimedes was quite put off by this excellent display of logic, so he decided to devise a better test. He told the triplets that he had eight stamps, four red and four green. Then he loosely affixed two to the forehead of each girl so that she could not see the stamps on her own forehead, but could see the foreheads of her sisters. He asked Kate, "Do you know the color of your stamps?" She replied, "No." The question was then put to Kathy and Katherine, with the same reply. Archimedes then asked Kate again, and she couldn't reply. Kathy, however, an89
swered, "Yes" on her second turn, and correctly named her stamps. What are the colors of her stamps, and how did she know?
Answers
1. Katherine reasoned thusly: "We all raised our hands to the first question, so there are at least two red hats among the three of us. If I am wearing a green hat, my sisters, much quicker than I, would have realized that each was raising her hand for the other. Therefore, I am wearing a red hat." 2. Archimedes explained the logic to me later. Kathy's correct answer depended on the fact that her sharpwitted sister Kate couldn't tell Archie her stamps on her second turn. Kathy probably started off by assuming that she had redred and then reconstructed Kate's thinking. Kate would have reasoned on her second turn as follows: "Ah, I see that Kathy has redred. If I also have redred, then all four reds would be used. Katherine, on her turn, would have realized that she must have two greens, and would have told Archie so. "But she didn't. That means that I can't have redred. Suppose that I have greengreen. In that case, Katherine again would have answered positively. She would have realized that if she also has redred, I would have seen four reds and would have answered that I have greengreen on my first turn. On the other hand, if she also has greengreen, then Kathy would have seen four greens and answered that she has two reds. Katherine would have realized that, if I have greengreen and Kathy has redred, and if neither of us answered on our first turn, she must have greenred. 90
EUREKA!
"But she didn't. That means that I cannot have greengreen either, and if I can't have redred or greengreen, I must have redgreen. And that means that I can tell Archie so." But she didn't, as Kathy saw. That meant that Kathy couldn't have redred, her original assumption. As Kate's argument also applied to Kathy's having greengreen, then Kathy couldn't have greengreen either. Therefore, as she answered, she must have redgreen. The reader could also have reasoned that, just as Kathy's logic applied to greengreen as well as redred, the solution is such that it is still true even if all the reds and greens are switched. This means that the person who answered "yes" must have had red and green stamps. Archimedes finally threw in the towel and gave full credit to the Pythagorean triplets for their skills in deduction.
91
11.
Better MixedUp Than Lost _ __
Sue Venir had just packed up six marblesthree black and three whiteinto three boxes labeled BB, WW, and BW, putting two marbles in each. Unfortunately, she realized that she had put each set of marbles into the wrong box, so that all the labels were incorrect. By picking just one marble from a single box (it would be such a bother to unpack them all), she managed to figure out which pair of marbles was in which box, and to adjust the labels accordingly. From which box did she pick?
Answer
She learned the contents of each box by drawing a single marble from the box labeled BW. If it had been black, then the other marble in that box must also have been black (or else the label would have been right). This means that the box labeled BB must have contained two whites, and the box labeled WW must have contained one of each color. Similar reasoning would apply if the first marble had been white.
92
EUREKA!
And Even @ Dissection ~oofSolids
12.
Archimedes Anderson and the Case of the Sinister Plot _ _ _ __
"What are you working on, Dad?" Archimedes Anderson had wandered into his father's study. "Anything interesting?" "Just a death out in the woods the editor asked me to write up." Mr. Anderson tipped back his chair and turned to face Archie. "A routine story, but there's something about it that doesn't seem quite right." He picked up a small notebook. "Sergeant Enigma lent me his notes on the case. Care to hear?" 93
Archimedes nodded, and Mr. Anderson assumed his best storyteller's manner .... The Sergeant took in the scene at a glance: a heavily wooded area, not lacking in logs or stumps, its tranquil beauty harshly broken by the body of Dan Druff, halfdried blood caking to the large depression where his right temple used to be. Half a foot away from his head, a red stake, obviously a boundary marker, poked out of the ground. "I'm sure that I can explain everything," a nervous Pete Moss said. "I've been planning to sell this corner of my property for a long time now. I knew Dan was a surveyor, so I asked him to measure the area for me." "You're selling off this land? Pity," Sergeant Enigma said, gazing up at the stately trees. "About what size plot?" "A triangular parcel about 850 feet by 650 feet by 1500 feet. I told this to Dan and he said that he wanted to look over the area before taking any measurements," Pete said, still a little jittery. "So, we were walking by this stake here, and Dan suddenly fell and caught his head on the marker. I tried to stop the bloodflow," he held up a soiled handkerchief, "but I couldn't. Soon he looked real dead, so I quickly went back to the house and called you." Sergeant Enigma determined that Dan Druff's surveying equipment was in its customary place in his car. Eventually he reported the death as an accident, and no arrests were made. Mr. Anderson finished and thanked Archimedes for his applause. "Something still seems a little fishy. As a writer, I realize that the circumstances are such that an accident would be a terrible waste of a great situation. But I'm afraid I don't know why."
94
EUREKA!
Archimedes thought for a moment and then suddenly snapped his fingers. "You're right, Dad! The catch is .... " How did Archimedes continue?
Answer
"The catch is that Pete Moss's measurements can't possibly be the sides of a triangle," Archimedes explained. "In a triangle, the sum of each two sides must be larger than the third. A triangle with sides of 850, 650, and 1500 would be nothing more than a straight line." "I see! Dan Druff, an experienced surveyor, would have realized that. He couldn't possibly have been walking on the grounds for that reason," Mr. Anderson continued. "Moss's story is a complete fabrication! Say, I'd better call Headquarters about this" Archimedes Anderson yawned and wandered out of his father's study.
95
13.
How to Dissect a Square and Other Marvels of Modern Biology _ _ __
"You like my table?" Archimedes Anderson looked rather pleased. "Great! I think it's rather neat myself." He cleared the chess pieces and schoolbooks off the square table. "Have you seen this yet?" With remarkable dexterity he folded the table out from the middle and swung the pieces around. Before Katherine Pythagoras's eyes the square became a perfect equilateral triangle! "This is one of my favorite geometric dissections," he said. She looked a little puzzled, and he added, "Geometric dissection is the changing of one polygon into another by cutting it up into little pieces and rearranging those pieces to form the new figure. It's quite a fascinating subject," he continued, rummaging around for a clean piece of paper. "This one is the 'haberdasher's problem' solved by one Henry Ernest Dudeney in England around the turn of the century. I especially like it because it can be done using only a straight edge and compass." Archie quickly drew an equilateral triangle ABC and bisected AB at D and BC at E. Then he extended AE to F so that EB equalled EF, and extended CB a little way past B. With the midpoint of AF as the center of a circle, he swung the compass from A through BC to F, calling the new point H. Now with E as the center of a circle, he drew arc HJ, with J a new point on AC. Then he put another point K on AC so that JK equalled BE. From D and K, perpendiculars were dropped onto EJ to obtain points L andM. 96
EUREKA!
"The final diagram looks like this," Archimedes said, pushing the paper over the table. (See Fig. 131.) He located a pair of scissors, cut the figure along the solid lines, and rearranged the pieces into a square (Fig. 132).
Fig. 131
Fig. 132
97
"Something particularly fascinating is that the four pieces can be joined like this to make a chain," he explained. (See Fig. 133.) "If closed one way it forms the square; closed the other way it makes the triangle. Dudeney made a brasshinged mahogany one, and that gave me the idea for the table. I had a terrible time placing the legs!"
Fig. 133
Dudeney's method, with modifications, can be adapted for nonequilateral triangles. A theorem first proved by the German mathematician David Hilbert, in fact, states that any polygon, no matter how absurd or unusual, can be transformed into any other polygon of equal area by cutting it into a finite number of pieces. Basically, this is done by cutting the polygons into triangles, dissecting the triangles into rectangles, and reversing the process to form the second polygon. Unfortunately, the theorem does not hold for solids whose faces are regular polygons, though some dissections of solids are known. Although the theorem shows that dissection of one polygon into another is always possible, the number of pieces may be very large. Henry Dudeney, accomplished in other areas of dissection, surpassed many longstanding records. For instance, Fig. 134 shows his elegant dissection of a pentagon into six pieces that form a square. 98
EUREKA!
,f. , I
, ,
",
"
,
,,'
"
,',I
,'1 I
,'
"
'" , '"
3
I
...
,I ' " .......
6 " 2 ............. " ........... ~,rJ_   ~,¥,,' '\.
"
\
1 ""\
"II.
4
"
..........•.
/ ,
.....••/
, 'v'
Fig. 13·4
A record that stood for ten centuries was the nine· piece dissection of three squares into one shown in Fig. 13·5 (dotted lines indicate extra cuts), found by the 10th century Persian astronomer Abul Wefa and broken by Dudeney with the ingenious six·piece solution shown in Fig. 13·6. (Draw circle with center at A. BC=DE=FG.)
.. I
I
I I I
'1I
; ~
I I
I
I
, I I
I
Fig. 13·5
99
5
6
1
D~'":""'~E,.IA''C
6
5
1
Fig. 13·6
Figures 137 through 139 show some more surprising dissections to contemplate. (The Greek cross, like the equilateral triangle, particularly lends itself to dissection.)
100
EUREKA!
Fig. 13·7 Hexagon to square (5 pieces)
Fig. 13·8
Dodecagon to Greek cross (6 pieces)
101
Fig. 139
Maltese cross to square (7 pieces)
1. If you were given Fig. 1310, could you reassemble it into an equilateral triangle? There are several ways of going about it. It is interesting to note that alternate solutions to a dissection are almost always completely unalike. Perhaps you can devise your own or a better one of these.
Fig. 1310
2. As a start, grind your teeth on the one in Fig. 1311, which is composed of five equal squares joined together. 102
EUREKA!
Can you, by making only two cuts (no tricks; just three pieces), rearrange it into a perfect square? The answers follow.
~
_ _......._ _....J _________ •
_____ .___ ..
~~....J
Fig. 13·11
Answers
1. Figure 1312 shows the easiest solution with those cuts.
Fig. 13·12
103
,, , \ \ \
1
\
Fig. 13·13
2. Figure 1313 is equivalent to solving the rather sticky dissection of five squares into one in only nine pieces. Congratulations!
104
EUREKA!
14.
Geometer's Heaven ________
1. Starting with a square of paper, Stan Doffish trimmed away the corners to leave the largest possible circle. Then he trimmed the circle to leave the largest possible square. How much of the original square did he discard?
2. Jim Nasium set up a ladder so that it rested on top of a 15foot fence and touched a building 8 feet away and some 27 feet up. How far from the fence was the lower end of the ladder? 3. If the circle in Fig. 141 has a radius of 10 units, how long is the diagonal x?
Fig. 141
105
Answers
1. The diagram in Fig. 142 should make everything clear. As the area of the inner square equals onehalf that of the outer, Stan snipped away half of the original paper.
Fig. 14·2
2. Again, a diagram (Fig. 143) is useful here. The proportion 15 is to 27 as x is to x + 8, gives x as 10, the distance from the end of the ladder to the fence.
106
EUREKA!
27
Fig. 14·3
3. In Fig. 144, line y is the radius of the circle, 10 units in length. It so happens that y is also the diagonal of the rectangle. Therefore, x is equal to y, so x is also 10 units long.
Fig. 144
107
15.
Hole in the Sphere _ _ _ _ _ __
Boris N. Ventive grinned fiendishly at his own cleverness. He quickly looked over both shoulders, and then pulled from under the hotel bed ... a bowling ball. Boris braced the bowling ball with one hand and reached for the drill with the other. He carefully cut out a solid plug through the center (leaving a large donut), put a small hollow cylinder inside the hole, and then glued the spherical cap from the plug onto the bottom of the lead container, creating a hollow cylinder inside an innocent bowling ball. See Fig. 151.
Fig. 151
Boris then removed a quantity of the world's most precious metal from its hiding place inside the televisionthe highly radioactive Tedium235. Being careful to touch the metal only with his gloves, Boris filled the lead container, and then glued the top spherical cap onto the top of the cylinder. His work was done. 108
EUREKA!
Done, but not unobserved. Special CIA agent Stu Pidd had seen the entire procedure through the keyhole. He desperately needed to know how much Tedium235 was being smuggled out of the country, but all he had to work with was the fact that he knew the length of the lead cylinder6 inches. He didn't know its width, but he thought that if he could figure out how much volume remained in the bowling ball, even though he didn't know its size, he could determine the volume of the metal. Stu wasn't able to, in any case. Using only the information he had, can you tell how much volume remained in the bowling ball after Boris had hollowed his 6inch cylinder through its center but before he had replaced the spherical caps?
Answer
The remaining volume was 361T cubic inches. The problem can be solved by reasoning that it has a unique solution, true no matter what size bowling ball is used or what the width of the cylinder is. In that case, the volume must be a constant that would hold even if the hole has a radius of 0 and a "length" of 6. That length would equal the diameter of the sphere. Therefore, the residue must equal the volume of a sphere with a diameter of 6 inches. The radius, then, is 3 inches. As the volume of a sphere is given by the formula V= (4/3)1Tr3, then (4/3)1T(3)3 equals 361T, giving 361T cubic inches as the remaining volume. So, for any size sphere, the internal (or "infernal") construction of a hollow cylinder with height 6 leaves a volume of 361T.
109
16.
Convexstasy _ _ _ _ _ _ _ _ __
If you wanted to stack four tennis balls so that each was
the same distance from all the others, you would find that you would have to put three down to form an equilateral triangle and place the last on top to form a pyramid. If you were then to connect the centers of these balls, you would get what is called a tetrahedron. The tetrahedron is a fourcornered solid with four equilateral triangles for the faces. It can be formed by folding up the three corners of an equilateral triangle so that they meet at a point. The tetrahedron is the first Platonic solid, or regular geometric solid. These figures have faces that are all the same regular polygon. The cube, next in the series, has six square faces, and the octahedron has eight triangular faces. It would seem that there would be an infinity of these Platonic solids, but there are only five! The three described above were known by the Egyptians, and the next two were discovered by the Greek mathematician Pythagoras. The fourth one, the dodecahedron, has twelve sides, all of them regular pentagons. It looks very much like a pointy soccer ball. The icosahedron, composed of twenty triangular sides, can be formed from the hexagonal shape by removing one of the triangular sections. The others fold down to form the icosahedron's corner. See Fig. 161.
110
EUREKA!
Fig. 161
QUICKIE
1. Which volume is larger? That of an icosahedron, or that of a dodecahedron that will fit in the same sphere? Of all these solids, you are probably most familiar with the cube, perhaps for its properties as a die. In the fantasy game Dungeons and Dragons, all these solids are used as dice. The octahedron, though, seems to be the most popular after the cube. If the octahedron is formed from the strip in Fig. 162, the opposite sides total to seven, just as they do on a cube.
Fig. 162
111
The same arrangement can be used to create quite an astounding trick. Ask a person to select a number from 0 to 7. Then hold up the octahedral die so that he sees the faces 1, 3, 5, 7, and inquire as to whether he sees his number or not. A "yes" answer has a value of 1. Ask about 2, 3, 6, 7, and add 2 for a "yes" answer; then ask about 4, 5,6, 7, and add 4 for a "yes." The sum is the number! 2. Perhaps the following problem is easier to figure out. Try to arrange the digits 0 to 7 on the faces of an octahedron so that the sum at each corner is 14. Good luck! Answers
1. You must have realized that the 12faced solid has a larger volume; otherwise, the problem isn't worth presenting. Most people figure that, because the icosahedron has more faces, it more closely approximates the sphere, and thus should have the larger volume. It doesn't. 2. There are three ways to arrange the digits; Fig. 163 illustrates one. By the way, this summing property is unique to the octahedron.
Fig. 163
112
EUREKA!
17.
Great Unsolved Problems    
There are three great unsolved problems of mathematics, ones which have stymied mathematicians for the past few millennia. The ancient Greeks could not figure out why they were impossible to solvebut perhaps you can.
1. The first involves duplication of a cube. Find the edge (exactly!) of a cube whose volume is twice that of another given cube. 2. The second concerns the trisection of an angle with the classic geometer's tools, the compass and straight edge. For any angle, find a method of cutting it into exact thirds. No fair using a ruler! 3. The last involves squaring a circle. Find a square whose area is precisely equal to that of a circle. If you find you can't do it, try circling a square, and if you give up, you'll find the answers following. Answers
1. The volume of a cube is equal to the edge cubed, or V = e 3 • Twice the volume of the cube is then equal to the cube of the cube root of two times the edge, or 2V = (e·~/2)3 . As the cube root of 2 is irrational (about 1.2599 ... ), doubling the volume of a cube (but not increasing it by 8 or 27 times) is impossible. 113
2. The second problem, trisection of a given angle, also involves cube roots, but in a much more hidden way. It is possible, using only a compass and straight edge, first to double an angle and then to "add" it on to the original, giving you three times your angle. Unfortunately, it is not possible to work backward. Also, the cube root is such that the trisection can be done if you mark a straight edge. Legal? It's debatable. 3. Squaring a circle is also downright impossible, because of pi's irrational nature. If you were to draw a circle with a radius of 1, the area would be exactly equal to pi (but already you've used a measuring device!). So, all you need to do is construct a square with an edge equal to the square root of pi. Alas, you cannot find the square root of an irrational, so this problem is sadly impossible. (Also, there is no equation that will enable you to draw a line exactly equal to pL)
114
EUREKA!
18.
Out of This World
Trying to comprehend having  8 apples or  3 dollars is like trying to understand the fourth dimension. There is a logical basis for both, but that doesn't make them any easier to understand. A point is said to have zero dimensions, neither length nor breadth nor height. If a point is moved, say, 1 inch, a onedimensional line with only a length is formed. If this line is moved 1 inch perpendicular to itselfso that its movement is at right angles to itselfa square is produced. The square has length and width, but no height. If many squares are "piled" on top of each other, or if this square is moved 1 inch perpendicular to itself by pulling it out of the paper, a cube is formed. The cube is threedimensional. Now comes the tricky part, the transition from the third dimension to the fourth. If this cube is moved in that inconceivable direction so that it is perpendicular to itself at all times, so that its movement forms right angles with all its dimensions, then it will be moving in the fourth dimension, not sideways, forward, or upward, but something else, a theoretical direction. To us, such a world in the fourth dimension is a world of superpowers. People in this extraspace can untie threedimensional knots without moving the ends. Just as we can touch the center of a circle or the heart of a twodimensional figure, so can a fourdimensional being perform surgery on our hearts without breaking the skin, or steal all the gold in Fort Knox by moving it in the fourth direction. 115
If a righthanded person is turned through 4space, he becomes lefthanded. The trouble is that he'll think everyone else has changed! A rubber band can be turned insideout in our space, but in the fourth dimension a basketball can be turned insideoutwithout tearing the sides. In our space, four tennis balls, piled so that each is touching and is equidistant from the other three, form a tetrahedron. If a fifth ball is to be added so that the distance between any two balls is the same, then it must be added in the fourth dimension to form an extratetrahedron. The most wellknown and basic fourdimensional shape, however, is the cuboid or tesseract. Just as a cube, viewed from above, can be seen as shown in Fig. 181, a tesseract can appear as the model (drawn in two dimensions) in Fig. 182. And just as the cube can be unfolded into a twodimensional cross (Fig. 183), the tesseract, bounded by eight equal cubes (they just look slanted) can be unfolded to form the threedimensional object shown in Fig. 184.
/
" V Fig. 181
116
EUREKA!
"
Fig. 182
Fig. 183
Fig. 184
A tesseract cannot exist in three dimensions. However, Robert Heinlein wrote about the woes of living in a cuboid in his delightful short story "And He Built a Crooked House." The story illustrates some odd pseudopowers of extradimensions; the characters see their own backs by looking into the next rooms. This is very hard to understand. It logically follows that 117
if there exist dimensions of 0, 1,2, and 3, there should also exist one of 4. But perhaps our problem in not being able to make the conceptual extension is that we can't imagine any space other than 3space. How can beings exist without height? Without width? We can't even construct a real twodimensional figureeven the printed letters you are reading have height. Perhaps the fourth dimension will turn out to be time or hyperspace, and not be geometrical at all. The question, though, will undoubtedly plague mathematicians for years to come.
118
EUREKA!
~ Photons Are Light B2)oMatter, Too
19.
Archimedes Anderson and the Gambling Candidate _ _ _ _ _ __
Archimedes Anderson was reading in the living room when he was interrupted by a peal of laughter coming from his father's study. Sighing, he put down his book and waited a few seconds. "Archie! You have to hear this," his father burst out from behind the closed door, shaking a sheaf of papers. "I was just writing up an article for the newspaper, and 1 came across an incredible piece of preposterousness. You want to hear?" Without waiting for an answer, Mr. Anderson settled 119
onto the sofa. "You've heard of Monty Carlo, haven't you?" Archimedes laughed. "Mostly I've heard about his ancestors and the great people he is descended from." "That's him exactly. Anyway, the members of the Society for the Aid to American War Veterans of Foreign Descent met Sunday to elect their new officers," he continued, glancing at his notes. "Monty Carlo insisted that he was the most qualified for the position of Chairman. To sum up quickly, he said that his forefathers had fought in the American Revolution (on both sides!) and the French Revolution, and that Carlos had fought in World War I and World War II." "Sounds like a good case to me," Archimedes agreed. "I'd elect him myself if I could." "He would have won, too, if he hadn't continued," his father said. "It seems he wanted to win next year's election at the same time. He went on to say that he was precisely onethird English, onethird French, and onethird German, and that his paternal grandparents were 'twice as English' as his maternal grandparents were French!" "Balderdash," Archimedes concurred, groaning. "If it weren't for the fact that ... " How did he continue?
Answer
"Balderdash," Archimedes concurred, groaning. "If it weren't for the fact that it's impossible, he wouldn't be a laughingstock today. Everybody has 21 parents, 22 grandparents, 2 3 greatgrandparents, and so on. Hence, Monty Carlo is saying that 2n is divisible by 3. Since 2 120
EUREKA!
is the only prime factor of 2 n , 3 cannot divide it evenly. Monty Carlo was a bit too eager to impress." "And the rest of his statement was garbage," Mr. Anderson nodded. "His gamble that nobody would realize this didn't payoff."
121
20.
Once Upon a Time . .. _ _ _ _ __
Age problems tend to be hard problems, and a lot of people would like to lay their hands on their inventors. 1. "The day before yesterday I was 13, but next year I'll be 16." On what day was this statement made? 2. I am 30 years younger than my father. If his age is placed after mine, a perfect square is formed. In 25 years, our ages will similarly form a perfect square. How old am I? 3. The ages of Jane and Mary added together make 44 years. Jane is twice as old as Mary was when Jane was half as old as Mary will be when Mary is three times as old as Jane was when Jane was three times as old as Mary. How old is Jane? 4. How old will you be when you figure that last one out? Answers
1. The person made the statement on January 1, and the person's birthday is December 31. 2. I am 14 years old. Both 1444 and 3969 are perfect squares (and their roots, 38 and 63, are 25 apart). 3. Jane is 27.5 years old and Mary is 16.5 years old. 122
EUREKA!
21.
A Problem Fly _ _ _ _ _ _ __
Abby Normal and Liz Ard are exactly onefourth of a mile apart and bicycling toward each other, Abby at 12 m.p.h. and Liz at 8 m.p.h. A tireless fly with a thirst for adventure begins flying back and forth between the two at a constant speed of 30 m.p.h., turning as soon as it reaches the bicycle of one (it started on Liz's bike, by the way). The fly continues this flight between the two, making shorter and shorter trips each time, until it meets a horrible death as the two cyclists bump front wheels. How far did the fly travel before the crash?
Answer
The fly traveled threeeighths of a mile. You may be surprised to find out that it doesn't matter whose bike the fly started on, or even that the speeds of the two cyclists were different. Liz and Abby together traveled 20 m.p.h. At that speed, it takes 45 seconds to cover onequarter of a mile. The fly, traveling at 30 m.p.h. during those 45 seconds, covered threeeighths of a mile before the terrible collision.
123
22.
The Leading Series of Pisa _ _ __
"Look, David, you've gotta get that book written soon." "I've started already. As a matter of fact, you interrupted me when you called" "Is that so? What are you working on now?" "An article on the Fibonacci series." "The what?" "Well, about 780 years ago a mathematician by the name of Leonardo of Pisa was working on a problem concerning the breeding of rabbits, and" "Rabbits! What have rabbits got to do with numbers?" "Fibonaccithat's Leonardo's nicknameasked himself how many pairs of rabbits there would be at the end of December if he started with one pair in January, if this pair produced one pair in February, if pairs always bred in the second month after birth, and if all the pairs bred, well, like rabbits." "Hairy problem. How'd he get the answer?" "By setting up a table of the number of rabbits each month. He found that there was an unusual sequence of numbers in the table, kind of an additive series." "Really? What's so special about it?" "The standard series begins with two is and each number after this is the sum of the two before it: 1, 1, 2, 3, 5, 8, 13,21,34,55,89, 144, 233, 377, and so on. By the way, 377 is the answer to the rabbit problem." "Interesting. What's it good for?" "The series? It's useful in many sciences, not only in mathematics. The spiral arrangement of buds on a plant 124
EUREKA!
can fit into a Fibonacci sequence, as can the number of bees in each generation of a male bee's ancestors. Electricians analyze electrical networks and scientists measure light reflection in Fibonacci terms." "You mentioned math tieins a minute ago." "And how! There are only two known perfect squares in the sequence, 1 and 144, while 1 and 8 are the only cubes. The 3d, 5th, 7th, 11th, and 13th numbers are all prime, while 3, 5, 7, 11, and 13 are also prime, and so on. Unfortunately, the 19th number is composite." "Slow down! If the nth Fibonacci number is prime, then so is n, but not always the other way around. Is that right?" "With the exception of 3, you've got it. There are also periodic factors in the series. Every 3d number can be divided by 2, every 4th by 3, every 5th by 5, every 6th by 8, and so on. And 2, 3, 5, 8 begin the series." "That makes sense. The same numbers are repeatedly added to form each new number." "That's a neat way of putting it. The Fibonacci series also ties in with the Golden Ratio, which is equal to (1 +y'5)/2, or about 1.618 followed by a string of decimals I've forgotten. Moving down the terms, 1/1 equals 1, smaller than the Golden Ratio; 2/1 equals 2, larger; 3/2 equals 1.5, smaller; 5/3 equals 1.66, larger; and so on." "You mean the ratio of successive terms approaches the Golden Ratio?" "Yes, as does any series formed the same way. It's the seesawing larger and smaller with the Golden Ratio, by the way, that makes an amazing paradox possible. If an 8 X 8 square with area 64 is cut into four pieces with various sides of 3, 5, and 8, the pieces can be rearranged to form a 5 X 13 rectangle." "It gained area!" "The paradox lies in the fact that a very small slice is
125
missing from the diagonal; 3, 5, 8, and 13 are all Fibonacci numbers. If x, y, and z are successive Fibonacci numbers, then xz  y 2 equals either 1 or 1." "I see. 52  (3 X 8) = 1." "Right on. If you added the 2 to those three numbers, you would have 52  3 2 = 8 X 2 = 16." "That's true for any consecutive four numbers?Gh, I've got to run now! Thanks for the talk. I hope you don't mind the phone bill we ran up." "What! You called collect!" "Click! "
126
EUREKA!
23.
The Early Something Catches the Whatever _ _ _ _ _ _ _ _ __
Justin Time lives in the town of Shapeless, 90 miles west of the city of Chris, Mass. One day he is given reliable information about the time of arrival of something at Chris. He estimates the speed of the something and figures out when it will arrive at Shapeless. But the something arrives sooner than he thinks because it traveled slower than he thought. Can you explain this strange event?
Answer
Justin Time had called the train terminal in Chris to find out when a certain eastbound train would arrive there (he did so because his local station was notoriously below standards in such cases). He was told that it would arrive promptly at noon. Justin decided that trains travel at about 90 m.p.h. In that case, as he estimated, the train would arrive at Shapeless at 11 A.M. But this train traveled at a constant 60 m.p.h., thus arriving in Shapeless at 10:30 A.M., or sooner than he expected.
127
(8)0 Shortcuts 24.
A Speedier System of Solving _ __
The Trachtenberg Speed System of Basic Mathematics was invented by the late Jakow Trachtenberg, a brilliant engineer and mathematician, during World War II. Imprisoned in a German concentration camp, he kept his sanity by devising new methods of calculation in his head, putting only the final formulas on hardtoget scraps of paper. Eventually, he escaped and founded the Mathematical Institute in Zurich, where his methods are successfully being used today. People the world over especially enjoy the new system because the only number written down is the answer!
129
Here are a few of the simpler methods for multiplication. Multiplication by 11: All you do is add the number to the digit on the right and carry any number in the lOs place. This is based on the fact that mUltiplying by 11 is the same as multiplying by 10, multiplying by 1, and adding the results. A tick mark (') indicates a carry of 1:
15637 X 11 = 172,007 17'2'0'07 The
7= 7+0
0=3+7
o=
6 + 3 + carry of 1 2 = 5 + 6 + carry 7 = 1 + 5 + carry
1=0+1 This method is especially useful when multiplying by large numbers. Multiplication by 12: This is a little harder. This time double each digit and then add the product to the digit on the right, as well as adding in any carryovers.
4372 X 12 = 52,464 5'2'4'64 The 2 in the answer, for example, is equal to (4 X 2) + 3 + carry of 1. Both of the above could have been done this way: Multiplication of twodigit numbers: Just set the two numbers down:
0023 X 14 130
EUREKA!
The first step is to multiply the righthand digits. Again, the' indicates a carry of 1:
..........
0023 X 14
'2 Now the multiplication is like so:
.......
0023 X 14 '2'2 The second 2 is the sum of (3 X 1) + (2 X 4) + carry of 1. The last step is to mUltiply the lefthand digits:

0023 X 14 = 322 3'2'2 The middle step can be repeated as many times as necessary, the two pairs of multiplicands creeping along leftward, until the last step is possible.
1. Try this one: 0065 X 38
DID YOU KNOW THAT ... 3025
=
(30 + 25)2?
And 2025
=
(20 + 25)2?
Squaring is another subject made easy by the Trachtenberg System. To square a number that ends in a 5, simply split it into two parts, the first being all the digits except the 5, and the second just the 5. Now mUltiply the first part by the number 1 more than it, and "join" it to 52 = 25. The square of 75 would look like this: 7 X 8 = 56, and 52 = 25; therefore, 75 2 = 5625.
131
This method also works with numbers of any size ending in 5, such as 765 or 9,876,234,105. For twodigit numbers beginning with 5, merely square the 5, add it to the unit's digit, and "join" it to the square of the unit's digit. The square of 54, for example, is 2916. The 29 = 52 + 4, and the 16 is the square of 4. Generally, the first step in squaring twodigit numbers is to square the righthand digit. For example, to find the square of 32:
The next step is to double the product of the digits and add any carry from the first step:
The last step is to square the lefthand digit and add any carryover:
32
2 =
10'24
1024
Twodigit squaring can also be accomplished by using the twodigit mUltiplication method outlined earlier. Adding long rows of numbers is another feature. You never count past 11! You start at the top of any column and keep a running total of the sum. As soon as it reaches 11 or more, simply make the tick mark and subtract 11 from the total. When you reach the bottom of the column, just write down the total and the number of ticks in that column below it. 132
EUREKA!
Here's an example: 758
492 1165' 129'1
49 + 122
2420 0133
running total ticks
Now add like this, in an L on the bottom two columns to obtain the final sum:
3 8 8 3,
the answer!
The 3 to the right = 0 + 3 8=2+3+3 8=4+1+3 3=2+0+1 Checking is also easy. 2. Try this one: 3689 758 9667 1064 6498 + 745
133
Answers
1.
2.
0065 24 67 40
X 38
3689 15'8' 9'661 1064 64'9'8' + 745 8858 1233 2'2'4'2'1 = 22 ,421
134
EUREKA!
=
2470
25.
A Letter Home _ _ _ _ _ _ _ __
Dear Dad, Here's a neat problem I dug up. I found the only answer myself, so you should be able to also. A tendigit number contains every digit from 0 to 9. The digits are arranged so that the number formed by the first two digitsreading from left to rightis divisible by 2, the number formed by the first three digits by 3, the first four by 4, and so on until the whole number is divisible by 10. What is the number? Love, David P.S. There are two digits which can be placed immediately. Which are they?
Answer
The number is 3,816,547,290, and the two digits referred to are the 5 and O. They can be placed immediately because a number divisible by 10 must end in 0, so the number divisible by 5 must end in 5. The easiest method of solution is to use divisibility rules to figure out the placement of sets of numbers. For example, every other digit, starting with the second,
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must be even; the sum of the first three digits must be divisible by 3; the third and fourth digits must make a number divisible by 4; the sixth, seventh, and eighth digits must make a number divisible by 8, and so on. Trial and error takes it from there.
136
EUREKA!
26.
In Which We Are Initiated Into the Secret Society of Square RootSolvers _____________________
Occasionally, in the course of doing a problem, you might find yourself in the rather unpleasant situation of having to find the square root of a number like 1625.9444 to three or four decimal places. With nary a calculator around, you're tempted to throw in the pencil. Don't give up hope! All you need to do is use the Algorithm Method of finding square roots (so called because the steps are repeated several times). DID YOU KNOW THAT ... The product of any four consecutive integers is always one less than a perfect square?
Step one: Just write your number under the radical sign, with as many pairs of numbers after the decimal point as you want places in your answer (actually, include one extra pair of Os for rounding). This example will show how to find the square root of 1625.9444 to three decimal places: V1625. 94 44 00 00 Now write the largest perfect square less than your number right underneath it, and put its root above: 40. V1625. 94 44 00 00 1600
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Already you have determined that the square root of 1625.9444 is between 40 and 41. Step two: Subtract and bring down the next pair of numbers. This is just like long division: 40. Y1625. 94 44 00 00 1600 125 94 Step three: Here's the tricky part. Double the quotient (really the approximation of the root so far) and put it to the side: 40. Y1625. 94 44 00 00 80 1600 I 25 94 Step four: Estimate the next part of the quotientbut there's a catch. The way to go about it is to estimate 80 into 259.4, or 800 into 2594. Either way you account for an extra place. The next part of the quotient is in this case 3 and a bit, so the digit 3 is tested. Put the 3 after the 80 and mUltiply it by 3. The product, 2409, is less than 2594, so 3 becomes the next part of the quotient. If the product were larger than 2549, then 802 X 2 would be the next trial. Step five: Now write the product down under the remainder from the subtraction and put the next part of the quotient above: 40. 3 Y1625. 94 44 00 00 1600 125 94 3 24 09 138
EUREKA!
Now subtract, as in step two, and bring down the next pair of digits: 40. 3 ...; 1625. 94 44 00 00 1600 94 3 24 09 [ 1 85 44
125
Step six: This is like step three, except that the new testingnumber can be found in two ways: You can either double the quotient to obtain 806, or add the 803 and 3 you used in the estimation. Step seven: This is just like step four. This time you estimate 806 into 1854.4, or 8060 into 18,544. The process continues in this loop indefinitely, until the number of places wanted is achieved.
The complete example is: 40. 3 2 3 0 = 40.3230 ... "';1625. 94 44 00 00 1600 [ 25 94 24 09 3 [1 85 44 2 1 61 24 [24 20 00 3 24 19 ;",2=9__ 171 00 0* 00 00 [71 00 ...
* The estimate was already larger than 7100 at this time, so another set of Os (nonexistent, actually; three places had already been achieved) was necessary. 139
QUICKIE Try this one to two places:
J477. 65 32 00 Got it? Check it in the Answer section. DID YOU KNOW THAT ...
Every odd number is the difference between two squares? The square root of n is the number of consecutive odd integers that can be subtracted from n?
Answer
8 5 5
140
EUREKA!
21. J477. 441 136 34 12 2
8 5 5 = 21.855 = 21.86 65 32 00 65 24 41 32 18 25 123 07 00 21 85 25 112175 ...
27.
Heads and Legs _ _ _ _ _ _ __
1. Herby Yore has a wellstocked barnyard. He keeps his pigs and chickens together; in the pen there are 13 heads and 36 legs. How many chickens does he have? 2. A traveling menagerie featured two kinds of sorry freaks of naturefourfooted birds with two heads and sixlegged calves. An attendant, having a little fun, mentioned to onlookers, "There are 18 heads and 100 legs together." He was terribly surprised when one of the audience told him how many birds there were. Can you do the same? 3. In a group of cats and canaries, the number of legs is 84 more than twice the number of heads. How many cats are there in the group?
Definition of Algebra: A hopeless search for the continually changing values of two mysterious letters, x and y. Answers 1. If x is the number of chickens, then 2x is the number of chickens' feet, 13  x the number of pigs, and 4(13  x) the number of pigs' feet. Then: 2x + 4( 13  x) = 36 2x + 52  4x = 36 2x = 16 x=8 Herby Yore has 8 chickens.
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2. If x is the number of birds and y the number of calves, then: y + 2x = 18 6y + 4x = 100 The solution of these equations (obtained by doubling the first to 2y + 4x = 36, and then subtracting the 36 from the 100 and the 4x from the 4x) gives 4y = 64, so there are 16 sixlegged calves and 1 twoheaded, fourfooted bird.
3. If x is the number of canaries and y the number of cats, then: 2x + 4y  84 2x + 4y  84 2y y
= 2(x + y) = 2x + 2y = 84 = 42
There are 42 cats and any number of canaries in the group; as a canary has twice as many legs as heads, the number of birds doesn't change the constant difference of 84.
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EUREKA!
28.
Noble Bases _ _ _ _ _ _ _ __
Representing numbers in bases other than base 10 is really much easier than it sounds. You just use powers of a number other than 10 to represent various values. A number like 576, equivalent to 576 10 , is equal to (5 X 10 2 ) + (7 X 10 1 ) + (6 X 10°). To change it into base 5, simply divide by the largest power of 5 less than 576, or 125. The quotient is 4, and the remainder is 76. The next lowest power of 5 is 25, and it goes into 763 times, and the remainder is 1. So, 576 10 is equal to 431 5, Likewise, 576 10 is also equal to 1001000000 2,484 11 , 12 574 , and 10 576 (oddly enough, N 10 = 10N ). Notice that the highest digit used in base N is N  1. Converting back is even easier. The number 1452 7 is equal to (1 X 7 3 ) + (4 X 7 2) + (5 X 71) + (2 X 7°). If the sum of these is found (343 + 196 + 35 + 2), 576 10 results. Similarly, 484 11 = (4 X 112) + (8 X 1P) + (4 X 11°) = 576 10 , Armed with this information, these trifling problems should be no trouble at all:
1. Mrs. Cosine has 43 5 students in her first class, 19 12 in her second, 11101 2 in the third, 417 in the fourth, and 1001 3 in her last class. How many students does she teach each day?
2. Jack Skidder's truck contains 440 5 cases. In each case, there are 33 7 boxes, and in each box there are 1100 2 shirts. How many shirts, in base 13 5 , is he transporting?
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DID YOU KNOW THAT ...
Answers
1. Mrs. Cosine daily teaches 23 + 21 + 29 + 29 + 28 = 130 pupils. 2. Jack Skidder's truck contains 120 X 24 X 12 34,560 10 = 103400 8 shirts.
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EUREKA!
=
29.
A Division in Ancient Rye _ _ __
Three travelers once met on the road at dusk and, as was the custom, prepared to share a campsite. One of the three, not having brought provisions, proposed that the other two should share their food with him, for which he would pay handsomely. This being agreed to, the second man produced 3 loaves of bread and the third 5 loaves, which they all shared equally. As payment, the first man laid down 8 pieces of silver for his share of the 8 loaves. The second man, who had provided 3 loaves, argued that he should get 3 of the 8 coins and the third man 5, but the third man thought he should have more. How should the coins be divided?
Answer
The three each ate 2~ loaves. The second man, then, gave i loaf to the first man, keeping 2~ loaves. The third man gave 2i loaves to the first, also keeping his share. Hence, 2i to i, or 7 to 1, is the ratio in which the money should be divided.
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30.
Getting at the Root of the Problem         
"Hello, is this Mr. Rusty Springs?" "Why, yes, it is." "Mr. Springs, this is Guy Dence from WALF. Guess what? You're on 'Are You aLert?' " "I am?" "Well, I hope you are, because you can win $1000 if you answer our special question correctly. Ready? Here goes, Rusty: Is the number 12,345 divisible by 3? Remember, you're on the air!" "Well, I, uh, urn, I don't know." "Did I hear you say no? So sorry, Mr. Springs, 12,345 is divisible by 3. Thanks for trying, from Guy Dence, WALF. So long .... " Had Rusty only known about digital roots, the prize would have been his. The digital root of a number is the onedigit sum of its digits. For example, 1 + 2 + 3 + 4 + 5 = 15, and 1 + 5 = 6, so the digital root of 12,345 is 6. It so happens that if the digital root of a number is divisible by 3, then so is the number. Any number with a digital root of 3, 6, or 9, such as 12,345, is divisible by 3. You may have noticed that the multiples of 918, 27, 36, 45, ... , 108, 117, and so onhave digits that add up to 9. In fact, if a number has a digital root of 9, then it is divisible by 9 (as well as 3). There are several very interesting things about 9 and digital roots, most of them arising from the fact that 9 is 1 less than 10, the base of the decimal system. In taking 146
EUREKA!
a digital root, any 9 can be cast out, so that the digital root of 12 is the same as the digital root of 192, 912, 2991, or 9,199,299. Also, the digital root of any number is the remainder when the number is divided by 9; 12,345 divided by 9 is 1371 with a remainder of 6. This works because the 2, for example, can be represented as either 2 X 1000 or 2 X 999 + 2. The 2 X 999 leaves no remainder, so the leftover 2 goes toward the digital root. Digital roots are not only useful in checking divisibility. They can also help determine if a large number is a perfect square, a perfect cube, or just plain perfect. All squares have digital roots of 1, 4,7, or 9, and cubes have digital roots of 1, 8, or 9. All perfects except 6 have digital roots of 1. If a number doesn't have the right digital root, then it doesn't need to be tested in a more complicated way. 1. Find the smallest number composed only of 1s and Os that is evenly divisible by both 3 and 25.
There is a particular dice game that uses digital roots. The two players agree upon an arbitrary large number, and the first player begins by rolling the die and scoring the number that is up. The second player turns the die a quarterturn and adds the number on top to the running total. The winner is the player who either lands on the goal or forces his opponent to go above it. The game is often played with 31 as the goal. The digital root of 31 is 4, so the first player can force a win only by rolling a 4 and either staying in the series 4132231 or preventing his opponent from doing so. The game is very hard to analyze because of the many possibilities; however, it is known that the second player can force a win if the digital root of the goal is 9. Otherwise, it's a dicey problem.
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2. Jack Daniels and Johnny Walker had gotten quite soused one wet Saturday night, and each was trying to prove that he was as sober as a judge but that the other was several sheets to the wind. "Oh, yeah?" Jack whined, disappointed over losing the most recent test. "That doesn't prove I'm drunker'n a skunk. But this'll show who is." "Hah! May my liver see the day!" Jack snatched up a deck of cards from two nearby gin players. With fingers made frisky from repeated applications of tonic, he layed out the clubs from the ace to the nine. "Here's a rum test I learned," he said. "I bet you can't do it. Arrange these cards to form a prime number." "That's no shaker! We royal Scots are prime mathematicians. " But hours later, he mumbled dispiritedly, "I give up," and, vaguely saying something about never having visited Tequila, died. He was placed on his bier the next morning. Johnny did have a tough problem, but this one is easier. Prove that it is impossible to scramble the first nine digits to form a prime. Answers
1. For a digital root so the base must end in
number to be divisible by 3, it must have a of 3,6, or 9. A digital root of 3 is the smallest, number is 111. For it to be divisible by 25, it 25, 50, 75, or 00. The number, then is 11,100.
2. The digital root of the sum of the digits from 1 to 9 is 9, so whatever number is formed will be divisible by 9; therefore it can't possibly be prime, or even perfect. It might, however, be a square or a cube. 148
EUREKA!
31.
Or is it 32? Remumber Nembers __________________________
We live in a world of numberszip codes, street numbers, dates, telephone numbers, Social Security numbersand I can't remember a single one. When I pick up the telephone, chances are that I'll dial someone in Cinome, N .M. or Davidle, Wis., instead of the person I want. All those long numbers just scramble themselves in my head. There is a solution! Memory experts have found that remembering creative words or phrases rather than numbers helps commit the numbers to memory. This chart has been developed to do just that:
0: Z or S 1: Tor D 2: N 3: M
4: 5: 6: 7:
R L J or CH or SH Kor C
8: F or V 9: P or B Let's say the number you want to remember is 43214, a zip code. Checking the chart, the 4s can be replaced by R, the 3 by M, the 2 by N, and the 1 by T or D. If you now insert some vowels into your word, you can have
REMAINDER. You can also break the number up into, say 4/32/14 if it doesn't work out so nicely. Then you can make the
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phrase ARE MEN DEAR by adding the vowels. This works quite well for long sets of numbers. A room number of 679 becomes .JACOB, so you can remember that Jacob lives in that room. Or perhaps he lives in apartment 32, on the MOON, or even in 820, on YENUS. There is even a mnemonic device (one of these convenient phrases) to help you remember the letters for the 10 digits in order: STNMRL(SH)CFP becomes ~ATAN MAY RELISH COFFEEEIE! These alphabetic mnemonics are really quite useful, but creating one for the number 1776 is an exercise for a true linguist. Another mnemonic device is to substitute digits with words, rather than letters. The length of the word is the digit. The 1776, then, can become I CREATED NOTHING SUNDAY. This technique has been used quite successfuly with two useful but hardtoremember constants, 7r and e. "How I wish I could recollect pi easily today" gives 3.14159265, and "Now I live a drear existence in ragged suits and cruel taxation suffering" carries pi out to 3.141592653589. The more absurd you make your own phrase, the better you'll remember it. The number e has several mnemonics: "To express e, remember to memorize a sentence to simplify this" gives e as 2.7182818284. "It enables a numskull to memorize a quantity of numerals" and "I'm forming a mnemonic to remember a function in analysis" are others. There was another mnemonic method, number three, but it's clean slipped my mind. I forgot the key word.
150
EUREKA!
7la Neat Numbers 32.
Prime Time _ _ _ _ _ _ _ _ __
The most studied class of numbers is the class of prime numbers. Primes are whole numbers which have no integer divisors other than themselves and 1. Examples are 2, 3, 5, 7, 11, 13, and so on, getting scarcer as the numbers get larger (1 is considered a prime only in England; this is oddit means that each prime is factorable into two primes, a statement which doesn't seem quite right according to our definition). It's wonderful to know that a number like 37 is prime, but who would believe that 9,090,909,090,909,090,909, 090,909,090,909,090,909,091 is prime? For that matter, consider even 909,090,909,090,909,090,909,090,909,091.
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How is it known that 1,234,567,891,234,567,891,234, 567,891 is prime? (Or so it is said.) Well, the first method of finding primes was called the "sieve of Eratosthenes." The person wanting to find primes would write down all the numbers from 2 to, say, 200. Then he would cross out all the multiples of 2 except for 2 itself. The next number not crossed out is 3, so all the mUltiples of 3 are crossed out. Four has been crossed out, so all the multiples of 5 are then crossed out, and so on. This continues until the numbers left are all primes. As you can see, this was a rather laborious and boring method of finding primes. Besides, people needed to determine if a specific number, such as 10,000,019, was prime. It was found that a number is prime if and only if it divided (n  1) ! + 1 evenly, where n is the number being tested. Still, finding 10,000,019! is, at best, unbearable. The most common and best method of determining primality goes right back to the definition; to determine if N is prime, simply divide N by all the primes less than ..[N. If a single one divides N evenly, then N is a composite number. If N were 10,000,019, this involves dividing 10,000,019 by the 446 primes less than 3162. Fortunately, this task readily adapts to modern electronic computers. (As of this writing, the first 6,000,000 primes have been compiled.) It sure would help if there were a formula to find primes. A lot of people thought they had found one, too. The polynomial x 2  3x + 43 gives primes for all values of x from 0 to 41, and x 2  x + 41 gives primes for x = 0 to 40. The simple 2x 2 + 29 generates primes for all values of x less than 29. Unfortunately, no polynomial (an equation of this type) can give wholly prime numbers. For those interested, the proof involves showing that the new function of x (2x 2 + 29, for example) eventually is divisible by x. 152
EUREKA!
The pattern: 3!2~+1!
4! 3!+2! 1! 5!  4! + 3!  2! + 1! seems always to result in a prime. Also, the product of all primes less than N, added to 1, also results in a prime for many values of N. This is called Fortune's Conjecture. The Tallman formula can boast of always giving a prime, but some manipulation is necessary. The formula finds the product of the first N primes, divides that by the product of any of those primes and/or unity, and then subtracts the second product. If the result, X, is less than the square of the (N + l)th prime, then X is prime. DID YOU KNOW THAT ... 19, 109, 1009, and 10,009 are all primes? Here is a polynomial that generates prime pairs, sets of primes that differ by 2, such as 3 and 5, 5 and 7, 11 and 13,19,469 and 19,471, and so on: N
=
60A 2

1710A + 12,150
where N + 1 and N  1 are the primes. This formula generates 18 pairs of positive prime pairs and 1 negative pair, 31 and 29, for A = 1 to 20. Certain primes, because of their nature, can be classified into certain groups. The numbers 11 and 1,111,111,111, 111,111,111, composed wholly of 1s, can be referred to as R2 and R 19 , where the subscript is the number of 1s. Other primes of this type are R 23 and R 3I7 , a huge number that would take up the rest of the paragraph. A different Rgroup, the Robinson primes, is made up of primes of
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the form 2 n k + 1, abbreviated R(k, n). The largest known Robinson prime is R(5, 1947), a number of 586 digits. Another class of primes is of the form 2 2n + 1; these are called Fermat primes. For n = 0 to 4, the Fermat primes 3, 5, 17, 257, and 65,537 are obtained, but F 5 is factorable. (Polygons withn sides, when n is a Fermat prime or the product of Fermat primes, can be drawn using only a compass and straight edge. The method was discovered by the famous Karl Friedrich Gauss in 1796, and is a procedure that eluded mathematicians for 2000 years. Gauss was not yet twenty at the time.) Oddly enough, F 7 was determined to be composite 65 years before its factors were found. F 8 is also known to be composite, but its prime factorization is not yet known. F 1945 is the largest Fermat number tested for primality, a gargantuan containing about 1010584 digits! It is also so large that its complete factorization is not known; however, one of its prime factors is the 586digit Robinson prime R(5, 1947). Now we come to the most famous class of prime numbers, the Mersenne primes, named after Father Marin Mersenne. In 1644, in the course of announcing the discovery of new perfect numbers, he stated that the form 2P  1 is prime for many prime values of p, such as 2, 3,5, 7, 13, 17, 19, 31, 61, 89, 107, ... , up to and including 11,213, 19,937, 21,701, and 23,209, the last values giving primes of 6000 digits or more. Testing these numbers for primality by dividing by all the primes less than Y223.20L 1, for instance, is ridiculous. In 1876, Edouard A. Lucas devised a special method of testing Mersenne numbers. He arranged a series 4,14, 194, 37,634, ... , where each term is 2 less than the square of the previous term. A number is prime if it divides the (p  1 )th term evenly. How it works: Let's suppose that we're testing 27  1 = 154
EUREKA!
127 for primality. (Notice that the three digits on the left are the same as the digits on the right. Also, 27 = 128; 1, 2, and 8 are all powers of 2. This is unique.) The first term of Lucas's series that is larger than 127 is 194, the third term. Dividing 194 by 127 leaves a remainder of 67. The fourth term is then 67 2  2 = 4487, which leaves a remainder of 42 when divided by 127. The fifth term of the series is then 422  2 = 1762, and the sixth, the (p  l)th term is then 1112  2 = 12,319, which leaves no remainder when divided by 127. Therefore, 127 is prime. Even this method is preferable to dividing large Mersenne numbers by thousands of primes. For every Mersenne prime, there corresponds a perfect number! (A perfect number is one whose factors add up to it.) Simply multiply 2P  1 by 2P 1, and a perfect number is sure to result. Furthermore, there is no known perfect number that does not correspond to a Mersenne prime. There are all sorts of interesting facts about primes of the form 4n + 1, such as 5, 13, 17, and so on. According to Fermat's Two Square Theorem, every prime of this type can be represented as the sum of two squares, as 13 = 4 + 9, but no prime of the form 4n  1 can be. Oddly enough, a prime of the form 4n + 1 is also once the hypotenuse of a right triangle, its square is twice, its cube thrice, and so on. Just how many primes are there? It has been proven that there is no largest prime, only a largest known prime (the proof involves showing that the product of any prime from 2 to m added to 1 either is prime itself or has a prime factor larger than m, as it can't be divided by any prime smaller than m). As mentioned before, primes become scarcer and scarcer, approaching a density of N / (In N), or N divided by the number that e must be raised to in order to get N. 155
There are many unsolved problems, conjectures, and theorems concerned with primes. Bertrand's Postulate, still not proven, states that there is at least one prime between nand 2n. Another unsolved problem: Is there always at least one prime between every set of two consecutive squares? [Hint: The density of perfect squares approaches 1!(2yN}] Goldbach's Theorem says that every even number is the sum of, at most, two primes. (So far, it has been proved that four primes will always suffice.) Does the theorem also apply to odd numbers? One odd pattern of primes is 333,337, 33,337, 3337, 337, 37, 7, in which each new prime is obtained by removing the lefthand digit of the previous prime. Aprimeprime is just the oppositeeach new prime is made from all but the righthand digits of the previous prime. One example is 317, which becomes 31 and 3; another is 31,379. Some others are 8 digits long! One last oddity: Primes can be in additive series, too. For instance, the series 5, 11, 17, 23, 29 has a common difference of 6. A longer chain of 10 terms starts with 199 and has a common difference of 210. A chain discovered in 1969 has 16 terms; it begins with 2,236,133,941 and has a common difference of 223,092, 870. You might want to search for series in which the difference between terms doubles, for instance.
1. Two women, mother and daughter, were celebrating their birthdays, which fell on the same day. "N ow, I remember," reminisced Geri Atrics, "that last year I was just twice your age." "That's true, Mom," Eva Porate agreed. "Have you noticed that both our ages are prime numbers now?" "Indeed, Eva," the elderly math teacher said. "What is more remarkable is that your age is the exact reverse of mine." 156
EUREKA!
What are their ages? 2. Frank A. Praisal, an assessor, was commenting on his income tax form. "I noticed that the ages of myself, my wife, and my son were all prime numbers and totaled 101. In 6 years they will all be primes again, and my wife's age and mine will total 100," marveled Frank. "How old is my son, and can he be listed as a dependent?" 3. The members of the Princesston University math team had finished their highstakes card game, and the debts were being settled. Jay Walker, the big loser, placed his check on the table and said, "There's a prime number of dollars for all of you." The other loser, who paid with four bills of different denominations, was surprised to find that he had also lost a prime number of dollars. "Well, I'll take a prime number and the rest of you can split up a prime number," Bob Sledd said as he picked up the check and laid down a hundreddollar bill. Four other winners each took a single bill, surprisingly always leaving a prime number of dollars, until the last winner took a prime number and left nothing. What is the smallest number of dollars that could have changed hands?
Answers
1. Geri Atrics is 73 and her daughter Eva is 37. 2. The son is now 13, Frank is 47 or 41, and Mrs. Praisal is 41 or 47. 3. The second loser paid with four bills, here called 157
a, b, c, and d, such that d is the largest and a the smallest, the sum of the bills is less than $100, and:
a is prime a + b is prime a + b + C is prime a + b + C + d is prime and a + b + C + d + 100 is prime As the three smallest denomination bills, 1, 2, and 5 dollars, add up to a composite number, c must be larger than 5 and must be a multiple of 10 dollars. Thus, the combinations of a and b are limited to 2 and 5 so that both a and a + b are prime. If a table is now set up with varying dollar amounts for c and d, three solutions are obtained. The smallest of these is: $227 + $37 = $264 lost; $2 + $5 + $10 + $20 + $100 + $127 = $264 won. A total of $264 changed hands.
158
EUREKA!
33.
A Sense of Balance _ _ _ _ _ __
1. Al Cohol, suffering from a latenight bout with the bottle, groggily noticed some unusual things about the kitchenwares he was transporting. He found that a jug, placed on the kitchen scale, balanced perfectly with a bottle. By experimenting a little, he also determined that a jug also balanced a cup and saucer, and that two bottles balanced three saucers. How many cups do you think it will take to balance a jug? 2. In a tug of war, four Briztles can pull as hard as five Gartxibles. One Briztle and two Gartxibles can hold their own against one mighty Dyquixt. If one Dyquixt and three Gartxibles waged a war of tugs with four Briztles, which side would win? 3. Miss Fortune was especially proud of the latest additions to her coin collectionnine valuable coins that looked exactly alike. "This one is worth the most," she said, removing it from its special case and comparing it with the others. "It weighs just a fraction of a gram more than the others." As she hefted it, the coin slipped from her hand and fell into the pile. "Oh, bother," she said. "Now I'll have to weigh them all to find the heaviest coin. That could take eight weighings." "Not so," disagreed Ty Kune. "I can find it for you in just two." How did he propose to do this?
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Answers
1. Three cups balance a jug. If we use the letters j, b, c, and s for jug, bottle, cup, and saucer, AI's first experiment showed thatj = b, or that
a jug weighs the same as a bottle. The second showed that j = c + s, which means also the b = c + s, as j = b. By multiplying by 3, the equation 3b = 3c + 3s is obtained. The last statement shows that 2b = 3s, so this is nothing more than 3b = 3c + 2b, by substitution. Then, b = 3c, and j = 3c. That is, a jug balances three cups. 2. In the last event, you can replace the Dyquixt with two Gartxibles and one Briztle. Then the tug is between four Briztles and five Gartxibles plus one Briztle. The first event showed that four Briztles tug as hard as five Gartxibles, so the team of three Gartxibles and one Dyquixt should win. 3. What Ty Kune proposes to do is weigh any three coins against any three others. If one set is heavier, he would weigh any two of the heavier set against each other; either one of those two is heavier, or, if they weigh the same, the third one is the heaviest. If the two sets balance, he would weigh any two of the other set against each other; again, either one of those two is heavier, or, if they balance, the third is the heaviest.
160
EUREKA!
34.
Perfect Numbers and Some NotSoPerfect Numbers _ _ _ __
Perfect numbers, in contrast to prime numbers, do have integer divisors. Euclid, founder of Euclidean Geometry, defined a perfect number as one which is equal to the sum of all its different divisors. The number itself is not included, or only 1 would be perfect, as it is its own divisor; for this reason, 1 is also not considered prime. The Greeks thought of the property of some numbers as "perfect." There is also a class of Multiperfect numbers, in which the sum of the divisors is a multiple of the number. For example, the sum of the factors of 120 equals 240 (first pointed out by Mersenne in 1631), and the sum of the factors of 672 is 1344, again twice it. Other numbers, considerably larger, have divisors adding up to 3, 4, 5, 6, 7, or 8 times the number. But back to mere perfects: For instance, 6 is equal to 1 + 2 + 3, and 28 = 1 + 2 + 4 + 7 + 14. These are the first two perfects. The next perfect numbers are 496, 8128, and 33,550,336, the fifth anonymously discovered around 1460. Perfect numbers are scarce, and so become large quickly. One of the last known perfects, the 24th, has 12,003 digits! You may have noticed that all these perfect numbers end in 6 or 8, and perhaps wondered if there are any odd perfects. Well, there are none smaller than 10 18 , but if they do exist, and there is some doubt that they do, they must be of the form 12m + 1 or 36m + 9, where m is a prime number.
161
Primes also play a part in the structure of even perfects. Euler proved in 1750 that all even perfects are of the form 2P  1 (2 P  1), where p and 2P  1 (a Mersenne number) are prime. If p = 2, the expression gives 6, a perfect, for instance, and 2 11 ,212 (2 11 ,213  1) is one of the larger perfects. Unfortunately, the number obtained is not perfect if p = 11; though 11 is prime, 211  1 is not. Just how many perfect numbers are there? We have seen that each Mersenne prime has a perfect counterpart, so determining how many primes of the form 2P  1 there are should tell you how many perfects there are. Still, proving that there is not an infinity of Mersenne primes (if there were, there would be at least as many, if not more, perfects) would not solve the problem of the number of perfect numbers until the question of odd perfects is resolved. Six is the "odd" perfect number. Besides the fact that the product of its factors is the same as their sum, 6 does not obey many of the rules that perfects seem to follow (also, 6 is the only number less than 10 that is not either a prime or the power of a prime). All known perfects, except 6, have digital roots of 1; that is, the ultimate sum of their digits is 1. For example, 4 + 9 + 6 = 19; 1 + 9 = 10; 1+0=1.
Every known perfect number, save 6, is also the sum of consecutive odd cubes, beginning with 1 and continuing until the number of cubes is equal to p=r. Must p be prime here? 28 = 1 3 + 496 = 1 3 + 8128 = 1 3 +
162
EUREKA!
33 3 3 + 53 + 3 3 + 53 +
73 7 3 + 9 3 + 11 3 + 13 3 + 15 3
Perfect numbers are also the sums of successive powers of 2 from 2P  1 to 2 2p  2 • This rule the number 6 obeys! 6 = 21 28 = 22 496 = 24 8128=2 6
+ 22 + 2 3 + 24 + 2 5 + 26 + 27 + 2 8 +27 +28 +29 +210 +211 +212
Again, must p and the number of terms be prime? Another "law" of perfect numbers that 6 obeys is that the sum of the reciprocals of its factors is 2. That is, t + ~ + 1 + = 2. This holds for all perfect numbers, and only for perfect numbers. Also, all perfect numbers, 6 included, are the sums of consecutive integers starting with 1. Amicable numbers are perfect numbers one step removed. If the members of such a pair are a and b, then the factors of a add up to b, and the factors of b add up to a. The bestknown and smallest pair is 220 and 284. Amicables differ from perfects in the fact that odd pairs, the smallest being 12,285 and 14,595, are known. All odd pairs are also divisible by 3, the lowest odd number other than 1, just as even amicable numbers are divisible by 2. Amicable numbers in a chain, in which the factors of one number equal the next, form what is called a sociable chain. One example is 12,496 + 14,288 + 15,472 + 14,264 + 12,496. (The arrows indicate that the factors of one number form the next number.) Chains of one link are perfect numbers, chains of two links are amicable pairs, and chains of four, five, nine, and more links are sociable chains. One mammoth chain, known for 60 years, has 28 links! Imagine figuring out that one without a computer!
i
163
~o 35.
Cranium Crackers and Cheese: Problems to Munch On
Classy Problems  _ _ _ _ _ __
There are quite a few ancient and decaying problems that are actually improved by twisting them around. Here is a small collection of the classic and modern versions of several rejuvenated antiques. la. Sam Ovar, the famous Russian tea magnate, was trying to figure out an efficient way of dividing his factories. He discovered that when he grouped them by 2s, 3s, 4s, or 5s, he always had 1 left over. What is the smallest number of factories he could have had before he sold that extra l?
165
lb. Several years later, a fortuneteller looked into her teacup long and hard and told Sam that he simply must reorganize his factories again. When he divided them into 2 groups, he had 1 left over again, but when he divided them into 3s, he had 2 remaining, by 4s he had 3 remaining, and by 5s he had 4 remaining. What is the smallest number of factories he could have had? 2a. Nancy France, away on a country vacation from her thriving resort hotel, came across an interesting problem. She, for some unknown reason, wished to cross a raging stream with a fox, a goose, and a sack of corn. The raft she wanted to use could carry only two things at once. If left alone, though, the fox would eat the goose or the goose would eat the corn. How on earth did she ever cross the creek? 2b. The fox in the above problem was none other than the Country Fox, Ursula Major. She owned the goose, of course. Nancy and Ursula found themselves in an awkward position. The raft would hold only two things, and the goose couldn't be left alone with the corn. To make matters worse, Ursula didn't trust Nancy alone with her goose and Nancy didn't trust Ursula alone with the corn. How did the four cross the river? 3a. Three tycoons, Smith, Brown, and Jones, live in Alfal, Fla. Three teachers with the same names as the tycoons live in the same town. Brown and the gym teacher live in the east end, Jones and the science teacher live in the west end, and Smith and the English teacher live in the south end. The gym teacher's namesake earns $100,000 a year, and the English teacher earns precisely onethird of the tycoon nearest him. If the teacher named Smith beats the science teacher at tiddlywinks, what is the English teacher's name? 166
EUREKA!
3b. Here's an odd bridge situation. Four players are named Arthur George, George Henry, Henry Thomas, and Thomas Arthur. Henry and George are playing against Henry and Arthur, and Thomas is sitting at Thomas's left. If partners sit opposite one another, who is Thomas Arthur's partner? 4a. Kim Ono had several robes which she distributed among her daughters. To the eldest she gave half the robes and half a robe. To the middle daughter she gave half of those left and half a robe, and to the youngest she gave half the robes and half a robe. Kim then had no robes left. How many did she start with? 4b. One day Kim decided to go wholesale. She packed up all the robes she had made into a truck and drove to a dealers' convention. The first customer bought half the robes and half a robe. The second and third customers did the same when their turns came. She then had just a quarter of a gross left. How many did she start with? 5a. The classic problem of the eight queens is easier to ask than to solve: How do you put eight pieces on an 8 X 8 chessboard so that no two pieces are on the same horizontal, vertical, or diagonal line? There are twelve solutions. 5b. The next step up is harder. Try placing sixteen pieces on the same board so that two and only two are on each row and column, and two at the most are on each diagonal. 6a. Henry Ernest Dudeney's best known puzzle is the problem of the spider and the fly. As shown in Fig. 351, the spider is one foot from the ceiling at the middle of the end wall of a room, and the fly is one foot from the floor at the opposite end of the room, perhaps caught in a web. 167
What is the shortest path the spider must crawl to reach the fly?
12
30 Fig. 35·1
6b. A less famous problem is this one. The fly is one inch down the inside of a cylinder four inches high and six inches in circumference. The spider is opposite, one inch from the bottom on the outside. How far must it crawl this time? See Fig. 352.
168
EUREKA!
Fig. 35·2
Answers
1a. The smallest number is 61. It is obtained by finding the smallest number all divide into. Because 4 has a factor of 2, the number is not 120 but 60. When the extra factory is added, 61 is the answer. lb. Each division falls just one short of being exact. The smallest number is then 60  1, or 59. 2a. Nancy took the goose across and left it there. She then took the corn across and took the goose back. She dropped the goose off and took the fox across. Then she went back and took the goose across, only to find that the fox had eaten the corn. 2b. Nancy first took the corn across and left it there. She went back and picked up Ursula, poled her across, and got off with her corn. Ursula went back and took her goose 169
across. Once there, they found the campsite they had been looking for and toasted each other's brains over roast goose and corn. 3a. The English teacher's name is Smith. From the last statement, we know that Smith is not the science teacher. One of the earlier statements says that the gym teacher's namesake earns $100,000 yearly. Because the English teacher earns onethird of this, the nearest tycoon is not named for the gym teacher. The English teacher lives in the same district as Smith, so Smith isn't the gym teacher. Therefore, he is the English teacher. 3b. From the second statement, the one about the pairings, it is clear that the two Henrys are opponents. Because the two Thomases are sitting next to each other, then two Henrys and two Thomases are opposed. Therefore, Henry Thomas and Arthur George are opposed by Thomas Arthur and George Henry. George Henry is the partner. 4a. Kim Ono started with seven robes. The problem is best solved by working backwards. To the youngest daughter she must have given one robe in order to have none left. In order to have one left, she must have had three before she gave robes to the middle daughter, and she must have had seven to start with. 4b. She started with 295. If she had 36 left after the third scale, she must have had 36 + 37 before it, and 73 + 74 before the second, and 147+ 148 = 295 at the start of the day.
170
EUREKA!
5a. Figure 353 shows what I feel is the most attractive of the twelve.
Fig. 353
5b. One solution is illustrated in Fig. 354.
Fig. 354
171
6a. Forty feet. The problem is best solved by unfolding the room. See Fig. 355.
/
/
;/
24
32
Fig. 35·5
6b. The same principle is used here. The minimum walk is five inches, as can be seen by using the unfolded cylinder (Fig. 356).
172
EUREKA!
4
6 Fig. 35·6
173
36.
LETTERS + DIGITS = FRUSTRATION _ _ _ _ _ __
Alphametics and cryptarithms fall into one of those categories of problems that you either love or hate. These ultimate word problems are gems in which a simple calculation such as addition or division is performed, but each digit in the equation is replaced by a different letter. The puzzle is to find the original equation; it's kind of like being given the answer and being asked to find the question. The "letterarithmetic" started thousands of years ago, but it was only in 1931 that the term "cryptarithm" was used for a puzzle like this one: ABC DE FEC DEC HGBC Not very appealing, is it? "Alphametic," a typo of alphabetic, was used in 1955 to represent puzzles in which phrases and words were used: BUT WE GET WET LOUT Believe it or not, this 174
EUREKA!
IS
the same puzzle. Much more
approachable, the challenge is still to find the original equationthis one is mUltiplication. This alphametic can be solved through the following steps (proceed with caution): 1. The multiplication of Wand B results in the W in the fourth line. Therefore, B equals l. 2. E and T are used a lot, and the product of E and T has an end digit of T (that is, T . E + T). The only values that work, Is and double digits excluded, are 0 . anything  0; 2 . 6 + 2; 4 . 6 + 4; 5 . 3 + 5; 5 . 9 + 5; and 8 . 6 + 8. Now, W . T + T also, so T must be 0 or 5 to have two values that result in T. If T = 0, then, because E . W + E in line 2, V would have to be 1. This can't be, because B is 1. Therefore, T = 5. 3. If T equals 5, then E and W must be chosen from 3, 7, and 9. Because W + 1 = L, a single digit, W cannot equal 9. 4. E cannot be 9, either. The product E . B does not result in E, so there must be a carryover. But 9 plus any carryover has two digits. Therefore, E doesn't equal 9. 5. If E = 3, then W = 7, because those are the only choices left. Because W . T + E, that would mean that 7 . 5 + 3. This isn't true, so E doesn't equal 3. 6. Therefore, E equals 7, and W equals 3. Because L = W + 1, L is 4. The sum E + T yields V, so V = 7 + 5 + 2. The product E· V, or 7 ·2, has a carry of 1, so G = E + 1 = 8. Then 0 equals 6. 7. Relax and enjoy it. There are more to come. The final equation is this: 125 X 37 875 375
4625
175
A couple of hints may help you solve the following problems. If N is an even digit, then N . 6 ~ 6. If N is odd, then N . 5 ~ 5. And if N . N = M, then M is 0, 1, 4, 5, 6, or 9. Good luck!
1. Here's a classic: SEND MORE MONEY 2. An easy warmup: LOSE SEAL SALES 3. Is this addition or subtraction? TRIED DRIVE RIVET 4. Two odd repetitions: a. ABCDE4 4
4ABCDE b. ABCDE 4
EDCBA
176
EUREKA!
5. A Joseph Trevor special. P stands for a prime number, either 2, 3, 5, or 7:
PPP PP
PPPP PPPP PPPPP Answers
1. M immediately equals 1, because it is a carryover. a equals zero, because the carryover cannot result in 11 or 12. S must be 8 or 9. If it is 8, then E = 9 because there must be a carryover to S + M. But then both N and a would be O. Therefore, S = 9. N + R > 10, so E + carry of 1 = N; and N = E + 1. E does not equal 2, because then N would be 3 and R would be 8 so that 3 + 8 + 1 ~ 2. Then D + E > 10 also. But D would have to be 8 or 9, and both digits would be used. E does not equal 3, because then N = 4 and R = 8. But D as 7 makes D + E = 7 + 3 ~ 0, already used. E does not equal 4, because R as 8 and D as 6 or 7 results in y as 0 or 1, both already used. If E = 5, then R can be 8, D can be 7, Y can be 2, and N can be 6. The solution: 9567 + 1085 10652
177
2. S must be 1 and A must be 0, as can be seen from the previous problem. S + A ~ E, so, because E can't be 1, E = 2. E + L ~ S, so 2 + L ~ 1, and L = 9. o + E ~ L, so 0 + 2 = 9, and 0 = 7. The solved problem: 9712 + 1209 10921
3. The problem can't be subtraction because T  D = R, so D< T. Then the E from E  V ~ E would have to be reduced by 1 to increase the D. V would be 9, but then so would I. So this is addition. From E + V ~ E, V equals either o or 9. If V = 0, then 1= 5, so R = 2 or 7. But T + D = R, so R could only be 7, making T + D = 6. But D + E ~ T and T > D (no carry, remember), so D = 2, T = 4, and E = 2. But then two digits are 2, so V can't be O. So V = 9. Then 1= 4, making R = 7. T + D = 6 still, but T < D this time, so T = 1, D = 5, and E = 6. The final equation: 17465 + 57496 74961 4a. Because 4· 4 ~ E, E = 6. D is found from E ·4+ 1 carry, so 25 ~ D, and D = 5. 4 . 5 + carry of 2 yields 22, so C = 2. 4 . 2 + carry of 2 yields 10, so B = O. 4 ·0+ carry of 1 yields 1, so A = 1. 178
EUREKA!
The final equation is: 102564
X
4
410256 4b. Because it is the unit's digit of a multiple of 4, A must be even. A must therefore be equal to 2, because A . 4 = E, a single digit. E is therefore 8. B must be 1 so that 4 . B is a single digit. D . 4 + carry of 3 from 32 ~ 1, so D = 2 or 7. A = 2 already, so D = 7. 4 . 1 + carry ~ 7, so carry equals 3. Then 4 . C has a lOs digit of 3, so C = 9. The solved puzzle: 21978
X
4

87912 5. This one is best solved by finding threedigit numbers, all of primes, that produce fourdigit numbers, all of primes, when multiplied by a prime. The only possibilities are: 775·3 = 555·5 = 755·5 = 325·7 =
2325 2775 3775 2275
The multiplier PP must consist of two identical digits, because there are no repetitions of any threedigit numbers.
179
There are only four possibilities to be tried, of which one works. The answer is: 775
X 33 2325 2325 25575 HOPE
YOU HADA + GOOD
TIME
180
EUREKA!
~o
37.
FUNdamental Ratios
Expand Your Mind _ _ _ _ _ __
Here are some absolutely unbelievable problems that are guaranteed to amaze and astound. 1. If you were to lengthen your belt by 2 inches, how much space would develop between the belt and your body? (Assume that you are a perfect sphere.) To be more realistic, if a cable 2 inches longer than the circumference of the earth were to be placed around the world, how much space would there be between the cable and the surface?
181
2. The milelong Frog's Neck Bridge, in Proge, N.Y., expands just 2 feet on a hot day. If there were no expansion joints to take up the extra length, how tall a bump would be formed in the bridge?
Answers
1. If D is the diameter of any circle, sphere, or midriff, then Drr is the circumference and Drr + 2 is the length of the belt or cable. The new diameter, then, is (Drr + 2)/rr, so the space on each side is equal to: [(Drr + 2)/rr]  D 2 which reduces to (2/rr)/ 2, or lIrr, about 0.3183099 ... inches, no matter what the original diameter! 2. If the bump were in the middle of the bridge, it would be the third side of a right triangle whose other two sides are ~ mile and ~ mile plus 1 foot (half the 2foot expansion). Then b 2 + 2640 2 = 2641 2 by the Pythagorean Theorem. This means that b 2 = 5281, or that the bump is nearly 73 feet tall! Drivers beware!
182
EUREKA!
38.
E? Ah! _ _ _ _ _ _ _ _ _ __
Let's just say you're out with a friend. In the course of casual conversation, your friend mentions that she has opened a new savings account at the bank. "It's for my oddjob money. Mom says saving it will be better than spending it all before prices rise." Halfinterested, you idly gaze at a hanging chain. "It's at only 4% interest a year, but the interest is compounded every week." "What?" You rise out of your stupor. "At the end of every week, the bank assesses the interest and adds it to my account. Then the next payment is interest on the interest." "Wow! How much do you have in there?" "Only a dollar so far," her eyes light up with expectation. "But you see it's going to grow very quickly!" You nod, but something doesn't seem quite right. And, little by little, it begins to dawn on you. Let's say your friend's bank didn't pay compound interest, but merely added 4 cents a year to her dollar. At the end of 10 years, her dollar would have grown to $1.40, and at the end of 25 years, the amount would have doubled. She would now possess 2 dollars. On the other hand, let's say her bank does pay compound interest. Just for the sake of argument, it compounds each account yearly. In 25 years, her dollar would grow to (1 + 1/25)25 dollars, or a little more than $2.66. Compounded every 6 months, the amount would grow to (1 + 1/50)50 dollars, or $2.69. But let's say the bank 183
compounds at the extraordinary rate of 52 times a year. You would expect the dollar to grow to quite a large sumif it were given enough time. And that's where she was caught. At the end of 25 years (or 1300 weeks), the dollar would have grown to (1 + 1/1300)1300 dollars,or just about $2.717, 2 1/2 cents more than what she would have made if the dollar had been compounded twice a year. And if the interest were to be compounded a million times a year, what kind of difference would that make? Your friend would earn a little more than onetenth of a cent over those 25 years than she would with weekly compounding. In fact, if the bank devoted every second of its time to compounding that lone dollar for the next 25 years, the amount of money at the end·would be bounded between (1 + 1/n)n and (1 + 1/nt+ 1, where n is twice the largest number you can think of. The amount would approach $2.718281828459 ... , a neverending, nonrepeating decimal. It doesn't matter if the bank pays 10%, 50%, or 172.6% interest. In the same time that the dollar takes to double at simple interest, at compound interest it approaches that mysterious 2.718281828 .... And that is the number e. Like pi, e is a transcendental number. This means that neither number can be expressed as the solution to any algebraic equation, nor be equal to any line segment drawn using the classic tools, a compass and straight edge. Thus e can be expressed only as an endless, continuing fraction or as the sum of an infinite series. For instance: e=2+1 1+1 2+2
3 + 3 ... 184
EUREKA!
(This continued fraction was discovered by Leonhard Euler. He was also the first to use the letter e for the number, and made so many discoveries about it that e is frequently called Euler's Number.) If the expression (1 + lin)" is expanded, one obtains the following series: e
=
1 + (1/1!) + (1/2!) + (1/3!) + (1/4!) ...
And eX = (xo 10!) + (Xl 11!) + (x 212!) + (x 313!) + . + (x" In!) .... By rearranging the terms, e can also be represented as (2/1!) + (4/3!) + (6/5!) + (817!) .... This little number pops up everywhere. Remember the hanging chain you idly eyed in the beginning of this Chapter? You got it. The curve, called a catenary curve, contains e in its equation. In growths of items in which the rate of growth is proportional to the size of the quan tity, e is also present; this is the case, in part, with the dollar and with the world population, for instance, among many other natural phenomena. The larger the snowball, the faster it gathers snow. 1. A problem comes to mind here. If a culture of bacteria doubles in population every hour and fills a test tube after 27 hours, at what point is the test tube precisely halffull? These processes are described by formulas based on y = eX; this is called the exponential function. In the world of calculus, e is its own derivative, and e, rather than 10, is the base of natural logarithms. Lastly, e is linked in this formula with 'IT and i, the imaginary square root of 1: e irr + 1 = 0, joining two transcendental numbers, unity, null, and an imaginary number in one equation.
185
2. Another little problem: Which is greater, err or rr e ? Try to solve it without the use of a calculator. Permutation involves factorial, so one would expect to find e involved in problems using permutation. Consider the following: 10 high school students leave 10 notebooks in a class. If the notebooks are later returned randomly to the students, what is the probability that no student receives the correct book? Firstly, there are 10!, or 3,628,000, possible arrangements of the 10 notebooks. It so happens that there are about 10!/e arrangements in which all the notebooks are wrong. This means that the probability of all the notebooks being wrong is (10!/e)10!, which is equal to lIe, and lie is about 0.3678794411 .... Thus, about 37 out of 100 times all the books will be returned to the wrong owner! Again, it should be no surprise to learn that the same number applies just as well to 10,000 students as to 10 billion. The probability is still about 3/8 that none of the participants will receive the right property. Since either all notebooks are wrong or at least one is correct, the probability of at least one student getting his or her notebook back is just about 5/8, no matter how many are involved. You might like to try this little stunt. Deal a shuffled deck of cards face up as you recite the 52 cards in some prearranged order. You win if you name the dealt card. Sound hard? Your chances of winning are l(l/e), or about 63/100. Try it on the nearest gullible millionaire. Memorizing e? You automatically know e to nine places if you know it to five, as four digits repeat (2.718281828 ... ). Otherwise, here are a few catchy mnemonics for e, in which the number of letters in successive words gives successive digits: 186
EUREKA!
"To express e, remember to memorize a sentence to simplify this." "It enables a numskull to memorize a quantity of numerals." "I'm forming a mnemonic to remember a function in analysis. " And, lastly, I'll leave you with two small problems. 3. What value of n gives the maximum value for the nth root of n? 4. To express e to six decimal places, there must be at least four digits in both the numerator and the denominator of the approximating fraction (as in 272111001). Can you find a threedigit fraction accurate to four places and a twodigit fraction accurate to three places?
Answers
1. The test tube was halffull at the 26th hour. 2. The greater is e1T , not
1fe.
3. The maximum value of the nth root of n is given bye. 4. The fractions are 878/323 (2.71826 ... ) and 87/32 (2.71875).
187
39.
A Section of Gold _ _ _ _ _ __
Phi is one of the more interesting irrational numbers (numbers whose decimal expansions are unending and nonrepeating). Not as well known as 1f or e, phi (symbolized by the Greek letter ¢) represents a fundamental ratio that pleasingly appears in many areas of mathematics.
A
B Fig. 391
The line in Fig. 391 has been divided into what is called the "Golden Ratio" or "Golden Section." The lengths of A and B are such that A + B is to A as A is to B. If B is equal to 1, the equation becomes: A + 1 =A

A

1
A+1=A2 0=A2_A1
Solved, this quadratic equation has the positive value of ¢, (1 + y'5 )/2, or 1.61803398 .... If A instead is 1 (by the way, if A = 1 mile, then B will be extremely close to 1 kilometer or about 0.62 miles, very close to 1!¢), the solution is equal to (y'5  1)/2, which, oddly enough, is equal 188
EUREKA!
to l/¢, or 0.61803398 .... Phi is the only positive number that becomes its reciprocal when 1 is subtracted (a negative number is l/¢). There's even a little more to this: ¢ + 1 = 2.61803398 ... , and ¢2 is also equal to 2.61803398 .... This series of 1, ¢, ¢ + 1 can be continued as 1, ¢, ¢ + 1, 2¢ + 1, 3¢ + 2, ... , where every term is the sum of the previous two. Amazingly, this series is equal to 1, ¢, ¢2, ¢3, ¢4, .... This is the only additive series in which the ratio between consecutive terms is constant. If a sq uare is to be cut so that the lengths of the pieces are in an additive series, such as 1, 1, 2, 3, 5, 8, 13, ... , and the pieces are to be assembled into a rectangle so that no area is lost or gained, the pieces must be cut according to the phi series shown above; otherwise, as the consecutive terms of any additive series have a ratio that approaches ¢, loss or gain of area must result. It's as easy as piein fact even easier than pito represent ¢ as the sum of an infinite series:
¢=V1+ V1+
vr+ V1+ VI + ... or:
¢=1+1 1+1 1+1 1+· ..
189
Phi can even be substituted for part of the second series. This yields = 1 + 1/ and 2 = + 1, which becomes the quadratic equation shown earlier. This proves that this series is correct (or that cJ> is eq ual to itself).
Fig. 39·2
Phi is also aesthetically pleasing. The Golden Rectangle (Fig. 392), whose sides are in the ratio of , has proven to be the one consistently used in everything from skyscrapers to doors to clothing to paintings. Leonardo da Vinci, for instance, applied the Golden Ratio to many of his works, connecting major figures in his paintings with the use of invisible Golden Rectangles. The ancient Greeks were well aware of . The Parthenon, a famous surviving temple, exhibits the Golden Ratio in its proportions. Phidias, a great sculptor of classical Greece, is believed to have used frequently in his work; perhaps he realized that the human body is divided into the Golden Section at the hip. Even the Egyptians were aware of . Recent measurements of the Great Pyramid of Giza (built about 3070 B.C.) reveal that the ratio of the length of the slant edge to 190
EUREKA!
the distance from the base center to the edge is almost exactly 1.618. Phi pops up in other places, too. It is, for instance, the ratio of the radius of a circle to the side of an inscribed decagon (a 10sided figure drawn so that the sides meet on the circle).
D
c
E
Fig. 393
Also, every segment in a pentagon is in Golden Ratio to the next smaller segment. (See Fig. 393.) One example is x/y; another is AC/AB. If ABC is included in the larger triangle in Fig. 394 (page 192) with a base of 1, then AC and AG are equal to (1 + V5)/2, phi. Also, BG is equal to (V5  1)/2, or 111)'>. The line AG is then divided into the Golden Section. 191
A
C~~G Fig. 394
D
A . ::::;;",_ _ _ _
L._ _ _L.......I
F Fig. 395
192
EUREKA!
C
The lengths of the triangle shown in Fig. 395 are 1,2, and yI5 (already you see what's coming?), and DE is 1. If AF is drawn equal to AE, the following results: AF AE AC FC = FC = AF = 1.618 ...
In the famous Fibonacci series, in which each term is the sum of the previous two (1, 1,2,3,5,8, 13, ... ), the ratio between consecutive terms approaches cpo For example 8/5 = 1.6; 21113 = 1.16154 ... ; 89/55 = 1.6182; 233/144 = 1.6181 .... 1. One last note: Can you show that x = cp if x(rx) = XX?
QUICKIE 2. An amateur mathematician had a circular pool, precisely 100 feet across, and two pool ladders. He decided to place them so that the shorter distance between them, measured along the edge, had the same ratio to the longer as the longer had to the entire distance. He had just placed the first ladder when Sir Cumference, an elderly geometry professor, arrived with a third ladder. "You mentioned to me what it was that you were doing," he explained, "and I realized that two ladders weren't enough for a pool this size. So, can you arrange the three so that the first distance is to the second as the second is to the first and second combined, and the second is to the third as the third is to both the second and third combined?" "But where will the third ladder go?" the owner wailed. "Well, I can tell you how far it is from the first," Sir Cumference said, "measured across the pool, but not from the second."
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What is the distance from the first to the third ladders?
Answers
1. If x(y1 x) = XX, then: X(Xllx)=xx Xl + I Ix = XX
1 + (l/x)
=
x
a= x2
 X 
1
=
(1 + V5 )/2
=
if>
2. The distance is 100 feet (See Fig. 396.) The problem is to set alb = bl(a + b) and blc = cl(c + b). As the same ratio pertains to both sets, alb = b/c.
3
c
2
1
Fig. 39·6
So, alb = blc and bl(a + b) = blc. From this, c = a + b, so c occupies half of the circumference of the pool, and therefore the third ladder is immediately across from the first. Actually, Sir Cumference didn't need to do this; he made use of the fact that in any series with the ratio of if> between the terms, each term is the sum of the previous two. So,c =a + b. 194
EUREKA!
40.
A BundledUp Buyer _ _ _ _ __
Greg Arious was accustomed to buying cylindrical bundles of asparagus 12 inches in circumference. One day his grocer instead packed two bundles of the same length, but only 6 inches in circumference, and insisted on charging the full price. Greg, on the other hand, was sure that the two bundles together contained less than one 12inch bundle. Who was right?
Answer
Greg was right. The circumference of a circle (the bundles were cylindrical, so a crosssection yields a circle) is equal to 27Tr, so the radii of the bundles are 6, 3, and 3 inches. The volume of the 12inch bundle is then 7Tr2 = 7T (6)2 = 367T, times the length of the cylinder; the volume of the two 6inch bundles is then 97T + 97T = 187T, times the length of the bundle. As the first figure is twice the second, Greg was right.
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41.
A Piece of Pi _ _ _ _ _ _ _ __
Few of us think of pi, or 1T, as representing anything other than the ratio of the circumference of any circle to its diameter. Actually, 1T is an integral part of mathematics and irrationally turns up in many areas. The ancient Egyptians, the first real mathematical culture, not only discovered the nature of 1T (the letter 1T is actually Greek), but also approximated a value for the number: (4/3)4, or 256/81, or 3.160+, really quite accurate. The fact that 1T, a decimal number, is irrational and does not repeat itself or end (as opposed to such a fraction as 9/11, which repeats and so is rational) was shown by the German mathematician Lambert in 1761. Pi was also proved by Lindemann to be transcendental. This means that it satisfies no equation with integer coefficients, such as 21T2  31T  2 = O. Other transcendental numbers are e, expressions involving natural logarithms (which are based on e), expressions involving trigonometric functions (which use 1T), and any number raised to an irrational power. Archimedes, the famous Greek philosopher and mathematician, went even further. He drew a circle with a diameter of 1, reasoning that a circle is actually a polygon with an infinite number of sides. This, joined with the alreadyknown fact that the circumference of a circle is equal to 1T mulitplied by the diameter, gives the circumference in this case to be equal to 1T. This means that, as polygons of greater and greater numbers of sides are drawn in the circle ("inscribed in the circle"), the perimeters of the polygons approach the circumference of the circle, 196
EUREKA!
and therefore get closer and closer to 71'. For instance, an equilateral triangle (Fig. 411) would have a perimeter of 3.67+, off by a little more than a half, and a hexagon (Fig. 412) would differ from 'IT by only 0.14+.
Fig. 41·1
Fig. 412
197
Archimedes eventually drew a polygon of 96 sides (try it yourself!) and found that 1f is between 3 117 and 3 1017l. The average of these two values is 3.141585 ... ,differing from 1f only in the tenthousandth place, quite a remarkable achievement. A circle with a radius of 1 has an area of 1f, and the square surrounding it ("circumscribed about it") has an area of 4. (See Fig. 413.) So, there is a probability of 1f/4 that a stone or coin dropped at random onto the square will also land in the circle. Experimentation can produce a very accurate value of 1f.
Fig. 41·3
In the same vein, if a set of parallel lines is drawn on a flat surface, perhaps 1 inch apart, and a Iinch needle is dropped onto the lines, the chance that the needle will fall across one of the lines is 21rr times the chance that it will not. This observation, made by Count Buffon, was tested in 1901 by the Italian mathematician Lazzarini, who 198
EUREKA!
obtained 3.1415929 ... after 3400 tosses. You might like to try one or more of these trials (yes, they are all trials!) and compare your value to 1f. Even before these tests were made, several actual formulas for 1f had been found. The first was discovered by Viete of France: 2/rr
=Vl72 X V 112 + 112y172 X
v'1I2 + 1I2V1I2 + 112y172 ...
James Gregory of Scotland found this one: 4/rr = 1 1 + 12

2 + 32 2 + 52
2 + 72 2 + ...
Pi is also equal to 4  (4/3) + (4/5)  (4/7) + (4/9) ... , and in slightly different form: 1f/2
2X2X4X4X6X6 ... 3X3X5X5X7 ...
= 
This is Leibniz's Formula. Another formula for 1f is 1f2/6 = 1 + (112)2 + (113)2 + (114)2 + (115)2 .... These formulas readily adapt to pocket calculators or computers. Before the electronic age, however, mathematicians did such calculations laboriously by hand, vainly searching for signs of repetition. In German texts, 1f is frequently called Ludolf's number, in honor of a 15th century mathematician who determined 1f to 35 places.
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William Shanks spent 15 years of his life calculating 71' to 707 places; his record stood until 1949, when a computer called ENIAC, after 70 hours of computation, spewed out some 2000 digits of 71'. (More recently, however, a computer calculated 71' to 1,000,000 places. The published result has been described as the world's most boring 400page book.) Alas, an error was found in Shanks's value, and the more than 100 digits following were all wrong! Memorizing 71' seems to be another favorite pastime among mathematicians. One Britisher memorized the number to some 5050 places (2217 repeats after 6 digits, 3.142857, and can thus be repeated until your tongue falls out; it is, at best, only an approximation of 71', differing after the second place). Various mnemonics, phrases or bits of prose in which the number of letters in successive words gives successive digits, have been devised over the years. "How I wish I could recollect pi easily today" gives 71' to 8 decimal places, and "May I have a large container of coffee" to 7. "Now I live a drear existence in ragged suits and cruel taxation suffering" gives a wonderful 12. This poem won a mnemonic contest set up by an English banking magazine, immortalizing 30 digits in prose: Now I will a rhyme construct, By chosen words the young instruct, Cunningly devised endeavour, Con it and remember ever Widths in circle here you see, Sketched out in strange obscurity.
200
EUREKA!
3.14159 265358 979 32384 626433 83279
A few more curious coincidences: 355/113 = 3.1415929 ... v'5T  4 = 3.1414+, 7r approximately 9/5 +V975= 3.1416+, again near 7r A great number of people do not enjoy the many intricacies and curiosities of 7r. For instance, the General Assembly of Indiana in 1897 enacted a bill to declare 7r de jure (according to law) equal to 4, for the most inaccurate version ever. In the same year, the Indiana State Legislature came within a single vote of declaring 7r to be 3.2. Actually, understanding pi is a piece of cake.
QUICKIE What is the radius of a circle whose circumference is equal to its area in terms of 7r?
Answer If the circumference of a circle is equal to its area, then 27rr = 7rr2. By factoring out a trr from each side of the equation, the expression r = 2 is obtained. The radius of such a circle must be 2.
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Selected Bibliography and Suggested Further Reading _ _ _ _ _ _ _ _ _ _ __
Andrews, W. S. Magic Squares and Cubes. New York: Dover, 1960. Barr, Stephen. Second Miscellany of Puzzles. London: CollierMacMillan, 1969. Cutler, Ann, and McShane, Rudolph. Trachtenberg Speed System of Basic Mathematics. New York: Doubleday, 1967. Dudeney, Henry. Canterbury Puzzles. New York: Dover, 1958. Fisher, John. Never Give a Sucker an Even Break. New York: Pantheon Books, 1976. Gardner, Martin. The Incredible Dr. Matrix. New York: Charles Scribner's Sons, 1976.   _ . The Unexpected Hanging and Other Mathematical Diversions. New York: Simon and Schuster, 1969. Hofstadter, Douglas R. "Metamagical Themas." Scientific American, March 1981. Lindgren, Harry. Recreational Problems in Geometric Dissection and How to Solve Them. New York: Dover, 1972. Lucas, Jerry. Championship Card Tricks. New York: Grosset and Dunlap, 1973. Manning, Henry P., ed. The Fourth Dimension Simply Explained. New York: Dover, 1960. Maxwell, E. A. Fallacies in Mathematics. London: Cambridge University Press, 1959. Scarne, John. Scarne on Teeko. New York: Crown Publishers, 1955. 202
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Schlossberg, Edwin, and Brockman, John. The Pocket Calculator Game Book. New York: William Morrow, 1975. Smullyan, Raymond. What Is The Name of This Book: The Riddle of Dracula and Other Logical Puzzles. Englewood Cliffs, New Jersey: PrenticeHall, 1978.
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