Matthew Wheeler Professor Winkler Math Puzzles 5/1/13 Simplifying Puzzles: Using Smaller Numbers Many math puzzles involve large amounts of numbers. The amount of numbers contributes to the difficulty of the problem. The Puzzle-maker knows that if the problem can be simplified to smaller numbers, the answer may be too apparent. For this reason, many puzzles involve a large amount of data to confound the puzzle-solver. The puzzle-solver can counteract the puzzle maker’s method however, by decreasing the numbers of the puzzle. This is my favorite method because it transforms difficult problems into much more manageable puzzles. Although the number-reducing method may not be applicable to every problem, it still is a powerful method for the problems that it can be used to solve. I was first introduced to this method when we discussed the checkerboard and domino puzzle in class. We then used this method to solve the five couples problem presented in class when it was suggested that we use the number reducing method. I was also able to apply this method to another problem I previously solved in Gardner’s book. I solved the problem much quicker when I applied the number-reducing method. The number-reducing method simplifies challenging puzzles and thus reveals the patterns that will be used to solve the problem. Before taking this class, I never considered reducing the numbers involved in a puzzle. Rather, I considered the puzzle to be concrete, and changing the parameters of it seemed like cheating at first. However, it is not cheating if you are able to observe the pattern involved in smaller numbers and then apply it to the original problem. It is essentially solving the puzzle, but with different numbers. This results in a much easier puzzle to solve. Once there, it is rather easy to work your way back up to the original puzzle while preserving the original relations. The mutilated chessboard puzzle asks the question of whether or not you can place 32 dominoes on a chessboard when you remove opposite corners of the chessboard. Intuitively, one would think that it is possible because each domino covers two spaces and, if two spaces are removed, the dominos should be able to cover the rest. Attemps at solving the problem reveal that it is not possible to cover all the spaces and that two spaces are always left uncovered. This leads to the eventual realization that since every domino covers a black and a white square independent of orientation. Thus, since the two squares that are removed are of the same color, the other squares will not be able to be covered. Rather than experimenting with the large 8x8 chessboard, it is easier to reduce the size right away. If we use a 4x4 chessboard, or even a 2x2 chessboard, we realize that it is impossible to cover the remaining squares much quicker than we do when trying to solve the original problem. The number reducing method makes this problem easier because it allows us to see the patterns underlying the problem without using numerous different domino orientations to attempt to solve the puzzle. This leads us to the solution quicker. The reason the number reducing method works for this particular puzzle is because the solution will always remain the same regardless of how many squares there are on the chessboard as long as the chessboard is bigger than 2x2. The dominoes will always have to cover a black and a white space, causing the chessboard to be any number of squares. The more squares, the more dominoes required to cover all the squares, however, two opposite colored pieces will always be uncovered because the opposite corners are the only two squares that are removed. Another puzzle that this method can be applied to is the Five Couples puzzle. In this puzzle, a wife and husband go to a party where there are five other couples. Various handshakes take place. No one shakes hands with himself/herself or with his/her spouse, and no one shakes hands with the same person more than once. After the handshakes are over, each person shook a different number of hands. The question is then how many hands the wife shook. The first step is to reason that the numbers of hands shaken are from zero to eight because no one can shake their spouse’s or their own hand, which reduces the number of possible handshakes from ten to eight. The next step is reasoning that the person who shook eight hands has to be married to the person who shook no hands because otherwise they only would shake seven hands. Following this logic, the solution to the puzzle becomes apparent The number reducing formula once again shortens the puzzle and makes it easier to solve. Rather than five couples, we use two couples. Thus, we only have four people involved in the new problem and three different possibilities for the number of hands shaken. With this method, it is clearer that, of the other couple, one must shake two hands and the other must shake zero hands because when they both give different numbers when asked. As a result of shaking two hands that must mean that the person who shook two hands shook both the wife and husband’s hand. This logic leads us to conclude that the wife shakes the same number of hands as the husband. If we increase the number of couples to three, the result is the same except for the fact that the wife and husband now shake two hands each. Following this pattern, it is evident that, for five couples, the wife and husband both shake four hands. The number reducing method works for the five couples puzzle because the pattern of handshakes is always the same. The wife and husband must always shake the same number of hands because there are only n-1 numbers available for handshakes and there are n people. So two people must have the same number. The husband has asked every one else what their numbers are and they all vary. This outcome leaves the wife with the same number as the husband. The last problem I realized this method would work on was the bills and two hats problem. A son asks his father for an allowance increase and the father presents a bet. The father has ten one-dollar bills and ten ten-dollar bills. The son can divide them any way he wishes into two hats. If the son draws a one-dollar bill, he loses; if he draws a ten-dollar bill, he wins. The problem is to figure out how the son should place the bills to maximize the chance of him drawing a ten-dollar bill. The trick to this problem is to realize that by placing one ten-dollar bill in one hat and the rest of the bills in the other hat, he maximizes his chances because he is guaranteed to win if he chooses the hat with the ten-dollar bill and he has nearly a fifty percent chance for the other hat. I figured this puzzle out after experimenting with the placements of the bills. For this problem, the number reducing method is once again applicable. If we have two bills of each dollar amount instead of ten, computing the probabilities becomes much easier and the problem is also easier to visualize. From this point, it is easier to reason that the best way is to place one ten into one hat and the rest of the bills into the other hat because there are less ways to change the placement of the bills. From this logic we can deduce that the best method for the four bills will work for any other number of bills. The number reducing method can be applied in this situation because the probability of drawing a ten for different methods will be related to the number of bills. The number reducing method is my favorite method because it makes seemingly complicated puzzles simpler and allows the puzzle solver to observe the patterns that determine the solutions of the puzzle. It also reduces the time elapsed to solve a puzzle because the puzzle-solver works with smaller number and detects the patterns quicker. Thus, the puzzle takes less time to solve.