MP CHAPTER 13 SOLUTIONS
MP SECTION 13.1 SOLUTIONS
1. If I can force it to be my opponent's turn with 1 match left, I will
win. Working backwards, if I can force my opponent's turn to occur with 6,
11, 16, 21, 26, 31 or 36 matches on the table I will win. Thus I should
pick up 40 36 = 4 matches on the first turn and on each successive turn
pick up (5  # of matches my opponent has picked up on his last turn).
2. Since the given information is symmetric (i.e. each player has won one
game) we may assume any order for the players' victories and get the
correct answer. Let's assume that Player 1 lost the first game, Player 2
lost the second game, and Player 3 lost the third game. Then working
backwards yields the following result:
Wealth
Player 1
Player 2
Player 3
After Game 3
10
10
10 (3 Lost)
After Game 2
5
5
20 (2 Lost)
After Game 1
2.5
17.5
10 (1 Lost)
Original
16.25
8.75
5
Note that we can't tell which player began with $16.25 or $8.75 or 5, but
we can be sure that 1 player began with $16.25, 1 player began with $8.75
and one player began with $5.
3. With one weighing we can find the heavy coin in a lot of three as
follows. Put any two coins on a scale (one on each side). If they balance
the third coin is the heavy coin; if they don't balance we can tell which
of the coins on the scale is heavier. Now we ask the following question:
if we had two weighings, for what number of coins could we find the heavy
coin? We must be able to get the set in which the heavy coin must be down
to three coins with one weighing left. We can easily do this if there are
7 coins. On the first weighing put three coins on each of the scales. If
the scales balance we have found the heavy coin; otherwise we have reduced
the problem to a 1 weighing problem for which the heavy coin must be in a
set of three coins. Thus if there are 7 coins we can solve the problem
with two weightings. This means that we can solve the problem with three
weighings if a single weighing can reduce the problem to one in which the
heavy coin must lie in a set of 7 coins. With 21 coins this is easy. Just
put 7 coins on each side of the scale. If the scales balance the heavy
coin is one of the 7 coins off the scale; otherwise the heavy coin is one
of the seven coins on the heavy side of the scale. Thus we can solve the
21 coin problem with three weightings. An illustration of the above
procedure follows:
First WeighingPut Coins 17 on left side of scale, 814 on right side of
scale, and coins 1521 off scale.
Possible Results of First Weighing
Left Side HeavierHeavy coin is one of coins 17
Right Side HeavierHeavy Coin is one of coins 814
Scales BalanceHeavy coin is one of coins 1521
Second WeighingLet's suppose we have found that the heavy coin is one of
coins 17. Then put coins 13 on left side of scale, coins 46 on right
1
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 Fall '07
 HOCHBAUM

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