Mathematical puzzles

Summary: The emphasis on problem solving in mathematics lends itself well to puzzles.

When considering mathematical puzzles, there are really two different types of puzzles available. Some puzzles are mathematical in nature, but require no mathematics to solve—similar to games like checkers, chess, and tic-tac-toe. Other puzzles are mathematical in nature and require a certain level of mathematics to solve—similar to games like cryptograms and Sudoku. Sometimes mathematical puzzles are referred to as “brainteasers.”

Tower of Hanoi

One of the oldest mathematical puzzles is the Tower of Hanoi. This puzzle was developed in 1883 by French mathematician Édouard Lucas. In the game, the player has several disks of different sizes and three pegs. The object is to move all of the disks from the starting peg to a different peg, according to the rule that a disc can only be placed on an empty peg or on top of a larger disc. In the legend believed to have inspired the game, there is a Vietnamese temple in Hanoi that contains a large room with three posts surrounded by 64 golden disks. The temple priests perpetually move the disks, according to the rules of the puzzle. According to the legend, when they are done, the world will end. If the legend were true, and if the priests moved disks at a rate of one per second, it would take them a minimum of 18,446,744,073,709,551,615 turns to finish—585 billion years. In general, the number of starting disks will determine the minimum number of moves to solve the puzzle.

To move a single disk requires only one move. To move two disks (D₁ and D₂ with the smaller number being the smaller, or topmost, disk) would require three moves: (1) D₁ to an empty, (2) D₂ to an empty, and (3) D₁ onto D₂. Three disks would require seven moves: move the top two disks as described above (three moves), move the last (bottom) disk to the empty, then move the two-disk stack onto the third disk (another three moves). A fourth disk would similarly require 7+1+7=15 moves. Using this pattern, the minimum number of moves for an additional disk will be double what the previous number of layers took plus one. However, to find the minimum number of moves for 10 disks, one needs to know what the minimum number of moves for nine disks would be. For nine disks, one needs to know the minimum number of moves for eight disks, and so on. Although a working recursive formula exists, it is not helpful for large numbers of disks. However, there is a pattern that can be found looking at the minimum number of moves for a certain number of disks that can be used to determine the minimum number of moves for any number of disks. In general, if there are n disks, the minimum number of moves to solve the tower problem will be 2ⁿ-1.

Two-Container Problem

Another old mathematics puzzle that was used in the 1995 movie Die Hard with a Vengeance involves two containers of different sizes that are used to measure a different third value. For example, in the movie, the characters were given a 5-gallon and a 3-gallon container and needed to measure exactly 4 gallons of water. It is assumed that there is an unlimited amount of water to pour into either container, and that contents of either container can be poured down a drain. Other versions of this puzzle can be formed by changing the size of the original containers or the quantity needed at the end. If the containers have capacities that are relatively prime to one another (greatest common factor is one), then any number less than the bigger container can be achieved. If the capacities are not relatively prime, then only certain values can be obtained. For this specific version, if x equals the number of times the 5-gallon container is filled and y equals the number of times the 3-gallon container is filled, the problem can be rewritten as an equation in two variables: 5x+3y=4.

Any ordered-pair solution to this equation will be a solution to the problem, although the method would still have to be determined. In the movie, the solution they found was (2, –2). The five-gallon bottle needed to be filled two times and the three-gallon bottle needed to be emptied twice (hence, the negative number). To actually solve the problem, they would have to fill the five-gallon container (first fill) and use it to fill the three-gallon container, leaving two gallons in the five-gallon container. The three-gallon container would then be emptied (first empty) and the remaining two gallons poured into the three-gallon container. The five-gallon container would then be filled again (second fill) and used to pour into the three-gallon container. Since the three-gallon container would have two gallons of water already inside, it would only hold one more gallon, leaving four gallons in the five-gallon container. The three-gallon container would then be emptied (second empty), leaving exactly four gallons. An alternate solution to this equation is (–1, 3).

Cabbage, Goat, Wolf

Another type of mathematical puzzle involves three objects and a keeper. As long as the keeper is present, all objects will remain safe, but if the keeper were to leave certain pairs of objects together unsupervised, at least one would be destroyed. For example, a farmer needs to transport cabbage, a goat, and a wolf across a river. The farmer is the only one who can row the boat and the boat is only large enough to carry the farmer and one other object. The goat and the cabbage cannot be left alone together as the goat would eat the cabbage. Similarly, the wolf and the goat cannot be left together as the wolf would eat the goat. The wolf has no interest in the cabbage, so that pair can be left alone together. The task is to determine how the farmer will get all three objects across the river.

On the initial row, the farmer’s only option is to take the goat. If he takes the cabbage, the goat is eaten. If he takes the wolf, the cabbage gets eaten. Once the goat is on the other side, the farmer leaves the goat and returns across the lake alone. The farmer must now choose to take either the cabbage or the wolf to the other side. The farmer returns to the first side with the goat and swaps the goat for the last object on the original side. Upon crossing the river, the farmer now leaves both the cabbage and the wolf on the opposite side of the river and returns to the original side with an empty boat in anticipation of picking up the goat. One final row allows the farmer and all three objects to be on the far side of the river.

Squaring a Double-Digit Number

Some mathematics puzzles take the form of mathematics magic. For example, if a spectator calls out any two-digit number, the mathematician can square the number without a calculator in a short amount of time—with practice, faster than a human verifying it on a calculator. Finding the square of some numbers is easy; for example, any multiple of 10 (such as 10, 20, or 30). All that is needed is to square the 10s digit and concatenate two zeros to the right. For instance, 70 squared would be 4900. A number that has a five in the ones digit is also easy to square; merely take the 10 digit, multiply it by the next-highest integer, and concatenate a 25 to the right. For example, to find 75 squared, take 7×8=56, then append 25 to get 5625. However, there are 90 possible two-digit numbers that could be called out and only 18 that fit one of the patterns above. For the remainder, the mathematician can employ a principle referred to as “squaring a binomial,” which is expressed algebraically as

(A + B)² = A² + 2AB².

If one needs to square a different two-digit number, such as 43, mentally rewrite 43 as (40+3). Using the above formula, the square can be found by

As mentioned above, 40 is a multiple of 10 and easy to square; similarly, 3 is easy to square. The more difficult part of the formula to calculate in one’s head is the middle—take 40 times 3 and double it. Then, add those three numbers together to get the square of the original number.

Squaring a number that has a 5 in the ones digit is a special case of squaring the binomial. If t equals the tens digit, then 10t+5 is the original number. Squaring the binomial yields

Factoring 100t from the first two terms yields

100t(t + 1 ) + 25.

Martin Gardner and Recreational Mathematics

Martin Gardner (1914–2010), an American mathematician, specialized in recreational mathematical games. From 1956 to 1981 he wrote Scientific American magazine’s Mathematical Games column and is credited by many for almost single-handedly sustaining and nurturing interest in recreational mathematics for much of the twentieth century. The kind of mathematical games Gardner wrote about are still being promoted not only for training children’s minds for mathematics, both in and out of school, but also for helping older citizens maintain sharp minds. In addition to paper and pencil books, there are many Web sites aimed at seniors that have mathematical puzzle collections, and popular handheld gaming devices (like the Nintendo DS) are now being targeting at consumers in all age groups for mathematics and memory games.

Bibliography

Behrends, Ehrhard. Five-Minute Mathematics. Providence, RI: American Mathematical Society, 2008.

Gardner, Martin. Hexaflexagons, Probability Paradoxes, and the Tower of Hanoi: Martin Gardner’s First Book of Mathematical Puzzles and Games. Cambridge, England: Cambridge University Press, 2008.

———. My Best Mathematical and Logic Puzzles. New York: Dover, 1994.

Vennebush, G. Patrick. Math Jokes 4 Mathy Folks. Brandon, OR: Robert Reed Publishers, 2010.

Winkler, Peter. Mathematical Puzzles: A Connoisseur’s Collection. Natick, MA: AK Peters, 2004.