# Matthew Wheeler Profes s or Winkler Math Puzzles 5/1/13

```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.
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