Math Trick For Amusements

1. How to form a loop with the Rolamite band while end A is taped to a tabletop is shown in Figure 109.

Figure 109

Figure 109

Solutions to the loop problem

Robert Neale, whom we encountered in the chapter on paper folding, suggested applying this to a playing card, say the joker. Use a razor blade to cut along the lines shown on the card at the left of Figure 110. Discard the shaded cut-out re-

Figure 110

Figure 110

The curious jokes

gion. By carefully executing the trick bend with the little square loop, taking care not to crease or tear its sides, you can produce the structure shown on the right. It is an amusing curiosity to carry in a wallet and show to friends. How the devil was it made? It looks, of course, as if the entire card had to be somehow pushed through the tiny window!

2. The first two rows of the chart [see Figure 111] show the results of the first two spins. The first spin was given, and we were told that D had the highest total after the second spin. This could happen only if the wheel distributed the points as shown in the second row. Now comes the tricky part. Every opposite pair of digits on the disk used in the game add to 5. This

Figure 111

spins

a

b

c

d

e

f

1

1

2

5

4

3

0

2

0

1

2

5

4

3

3

5

4

3

0

1

2

4

5

4

3

0

1

2

5

4

3

0

1

2

5

final scores

15

14

13

10

11

.12

Solution to D. St. P. Barnard's problem

Solution to D. St. P. Barnard's problem means that every spin will give a combined sum of five points to each pair of players seated opposite each other—namely AD, BE and CF. At the end of the game, which has five spins, each of these pairs of players will have a combined sum of 25 points.

We know that A won the game. Since his was the highest score, D (who sits opposite) must have ended with the lowest score. D's final score must be less than 13, otherwise A's final score would be smaller. D's final score cannot be 12. True, A would score 13, but then a player of the pair BE, as well as a player of the pair CF, would necessarily score 13 or better, preventing A from being the highest scorer.

As we have seen, D cannot score more than 11. He already has nine points at the end of the second spin, threfore, at least one of the three remaining spins must give him zero. Since the order of the results of each spin cannot affect the final scores, we can assume that D scored zero after the third spin. This determines the points for the other players as indicated in the third row of the chart.

On the next two spins, D's points can only be 0—0, 0—1, 1—1 or 0—2. We test each in turn. If 0—0 or 0—2, A will tie with someone on his final score. If 1—1, F gets 5—5 and wins with a score of 15. Only 0—1 remains for D. This makes A the winner, with 15 points, and enables us to complete the chart as indicated. We do not know the order of the last three spins, but the final scores are accurate. The problem is No. 2 in D. St. P. Barnard's first puzzle book, Fifty Observer Brain-Twisters (Faber and Faber, Ltd., 1962).

3. The formation rule for H. S. M. Coxeter's frieze patterns is that every four adjacent numbers b a d c satisfy the equation ad = be + 1.

4. Walter van B. Roberts answered his beer-can problem this way: "Imagine that the beer is frozen so that the can of beer can be placed horizontally on a knife-edge pivot and balanced with the can's top to the left. If it balances with the pivot under the beer-filled part, adding more beer would make the can tip to the left, whereas removing beer would make it tip to the right. If it balances with the pivot under the empty part, the reverse would be true. But if it balances with the pivot exactly under the beer's surface, any change in the amount of beer will make the can tip to the left [see Figure 112]. Since in this case the center of gravity moves toward the can's top when any a

BEER

BEER

pivot w pivot

The balanced beer can change is made in the amount of beer, the center of gravity must be at its lowest point when it coincides with the beer's surface.

"With the can balanced in this condition, imagine that the ends are removed and their mass distributed over the side of the can. This cannot upset the balance because it does not shift the center of gravity of the system, but it allows us to consider the can as an open-ended pipe whose mass per unit length on the empty (left) side is proportional to the weight of an empty can, whereas the mass on the beer-filled right side is proportional to the weight of a full can. The moment of force on the left is therefore proportional to the weight of an empty can multiplied by the square of the length of the empty left side, and the moment on the right side is similarly proportional to the weight of a full can multiplied by the square of the length of the beer-filled right side. Since the can is balanced, these moments must be equal.

"Pencil and paper are now hardly required to deduce that the square of the length of the empty part divided by the square of the length of the full part equals the weight of a full can divided by the weight of an empty can, or, finally, that the ratio of the length of the empty part to the full part is the square root of the ratio of the weight of a full can to an empty one."

Expressed algebraically, let a and b stand for the lengths of the empty and filled parts of the can when the center of gravity is at its lowest point and E and F for the can's weight when empty and full. Then a2E = b2F, or

In the example given, the can weighs nine times as much when it is full as it does when it is empty. Therefore the center of gravity reaches its lowest point when the empty part is three times the length of the full part, in other words, when the beer fills the can's lower fourth. Since the can is eight inches high, the level of the beer is 8/4 = 2 inches.

After the above solution appeared in Scientific American, Mark H. Johnson wrote to say that the answer is not strictly accurate. Because the tops and bottoms of the frozen can are at unequal distances from the pivot, they exert unequal moments of force. Distributing their masses over the can's side, to make a uniform and open pipe, would tilt the can slightly to the air side. To solve the problem precisely one needs more data about the can's dimensions and the masses of its top, bottom and side. Other readers reported that the solution also neglects what naval architects and engineers call the "free surface effect." When liquids are free to move inside containers, a slight raising of the vessel's center of gravity results.

5. Regardless of which coin Smith chooses, the game is fair. The payoff matrices show [see Figure 113] that in every case the person least likely to win (because he has only one coin) wins just enough when he does win to make both his expectation and that of his opponent zero.

As David Silverman suspected when he found this solution, the problem is a special case of the following generalization. If

Figure 113

75

50

25

0

win

75

75

75

75

sum = 300

loss

-100

-100

-100

0

sum = 300

125

100

25

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win

-50

-50

125

125

sum = 150

loss

-50

-50

-50

0

sum = 150

150

100

50

0

win

-25

-25

-25

150

sum = 75

loss

-25

-25

-25

0

sum = 75

Payoffs for player with silver dollar (top,) half-dollar (middle), and quarter (bottom)

a set of coins have values that are adjacent in the doubling series 1—2—4—8—16 . . . and the game is played as described, it is a fair game regardless of how the coins are divided. We assume, of course, that each player has at least one coin and that each value is represented by only one coin.

Daniel S. Fisher, a high school student in Ithaca, N.Y., generalized Silverman's generalization. He showed that Silverman's game is fair for any division of ownership of the coins when values of the coins are 1, n, n2, . . . , nk and the coins are weighted to fall tails with probability 1 In (n equal to or greater than 2).

6. The maximum number of nonoverlapping triangles that can be produced by seven, eight and nine lines are 11, 15 and 21 respectively [see Figure 114]. These are thought, although not yet proved, to be maximal solutions.

Figure 114

Figure 114

7. The solution with the largest product is:

532 98

7,448 7,448

The problem has 11 basic solutions:

532 x 14 = 98 x 76 = 7,448 584 x 12 = 96x73 = 7,008 174x32 = 96x58 = 5,568 158x32 = 79x64 = 5,056 186x27 = 93x54 = 5,022 259 x 18 = 74x63 = 4,662 146x29 = 73x58 = 4,234 174x23 = 69x58 = 4,002 134x29 = 67x58 = 3,886 138x27 = 69x54 = 3,726 158x23 = 79x46 = 3,634

Many readers found all eleven by hand, others found them with computer programs. Allan L. Sluizer pointed out that the maximum answer has digits 1 through 5 in one of the multiplications, and digits 6 through 9 in the other.

If 0 is included among the digits (though not as an initial digit of a number), we may ask for solutions of the expression abc x de =fgh x ij. There are 64 solutions, all independently found by Richard Hendrickson, R. F. Forker, and Sluizer. The one with the smallest product is 306x27 = 459x 18 = 8,262. The one with the largest product is 915x64 = 732x80 = 58,560. The maximum solution is given by Dudeney, in his answer to problem 82 of Amusements in Mathematics, as the maximum product obtainable if the ten digits are divided in any manner whatever to form a pair of multiplications, each of which gives the product. As in all such problems, 0 may not be an initial digit. The lowest product is given by 3,485 X 2 = 6,970 x 1=6,970.

"It is extraordinary," Dudeney once declared, "what a large number of good puzzles can be made out of the ten digits." Here are some examples similar to our original problem. How many solutions are there to ab X cde = fghi, using the nine positive digits? And how many to axbcde=fghi7 The seven solutions to the first problem, and the two to the second, are given by Dudeney in his answer to problem 80, Amusements in Mathematics.

Using all ten digits, how many solutions are there for abx cde=fghij? I have not seen this answered in print, but Y. K. Bhat, a correspondent in New Delhi, found nine:

39x402= 15,678 27x594= 16,038 54x297= 16,038 36x495= 17,820 45x396= 17,820 52x367= 19,084 78x345 = 26,910 46x715 = 32,890 63x927 = 58,401

How about ab~Xc = de+fg = hi, excluding 0? In Modern Puzzles, problem 73, Dudeney gives the only answer: 17x4 = 93 + 25 = 68.

Clement Wood, in his rare Book of Mathematical Oddities (Little Blue Book No. 1210), asserts that abxc = dexf=ghi (0 excluded) has only two solutions: 38x4 = 78x2=156, and 58x 3 = 29x6= 174.

One final problem that I leave unanswered. Find the only solution (excluding 0) to a x be = d x ef= g x hi. This was sent to me in 1972 by Guy J. Crocker, who discovered it. I cannot recall having seen it before.

8. If n is odd, it is obvious there is no solution. If n is even but not a multiple of 4, an odd number of checkers must be jumped on the final move. This would necessarily leave a single checker in the row, therefore the assumption that there is a solution when n is not a multiple of 4 must be false.

If there are 4n checkers, the problem can be solved by working it backward according to the following algorithm. Start with n/2 kings in a row. Take the top checker from either of the two middle kings, jump over the largest group of kings and put down the checker as a single man. On the next backward move take the top checker from the other middle king and jump in the same direction as before, jumping one fewer checker. Follow this procedure until all kings in the direction of the first jump are eliminated. Take the top checker from the inside king and jump in the same direction as the previous jumps, moving it over the proper number of checkers. Continue this procedure, always in the same direction, until all the kings are reduced to single men. When these moves are taken in reverse order, they provide one solution (there are many others) to the original problem.

9. Here are six possible explanations of the ski tracks:

(1) The skier bumped into the tree but protected himself with his hands. Keeping one ski in place, he carefully lifted his other foot and moved to the lower side of the tree. With his back against the tree, he replaced his raised foot and ski on the other side, then continued down the slope.

(2) The skier slammed into the tree with such force that his skis came off and continued down the slope without him.

(3) Two skiers went down the hill, each wearing only one ski.

(4) One skier went down the hill twice, each time with one ski on one foot.

(5) A skier went down a treeless slope, moving his legs apart at one spot. Shortly thereafter a tree with a sharpened trunk base was plunged into the snow at that spot.

(6) The skier wore stilts that were high enough and sufficiently bowed to allow him to pass completely over the tree.

So many readers sent other preposterous explanations that I can give only a sampling:

A small, supple tree that bent as the skier went over it was proposed by John Ferguson, John Ritter, Brad Schaefer, Oliver G. Selfridge and James Weaver. Ferguson also suggested (among his 23 possibilities) a pair of skis pulled uphill by long ropes and two toboggan teams of very small midgets, four on each ski. Selfridge included this one: The skier, aware of his ineptness, wore a protective lead suit. His impact on the tree sheared out a cylindrical section. The dazed skier passed between top and bottom parts of the tree before the top fell down and balanced perfectly on the base.

Manfred R. Schroeder, director of the Drittes Physikalisches Institut at the University of Gottingen, reported an actual experience he had in 1955 while skiing down a mountain in New Hampshire. "I hit a small but sturdy tree with my right shin-bone. The binding came loose and the ski and leg went around different sides of the tree. Below the tree, leg and ski came together again. However, the binding did not engage (no automatic step-in bindings then!) and the tracks ended in a spill about ten yards farther down the slope. Even then, in spite of considerable pain in the leg, I thought it was a worthwhile experience."

Johnny Hart, in his B.C. comic strip, has played with the theme. Thor, speeding toward a tree on his stone wheel uni-cycle, once went around the tree leaving two tracks. What happened on a later occasion is reproduced in Figure 115.

BIBLIOGRAPHY

For more on the beautiful properties of frieze patterns see:

"Triangulated Polygons and Frieze Patterns." J. H. Conway and H. S. M. Coxeter. Mathematical Gazette, Vol. 57, October 1973, pages 87-94.

"Additive Frieze Patterns and Multiplication Tables." G. C. Shephard. Mathematical Gazette, Vol. 60, October 1976, pages 178-184.

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