# CH13 - 1 MP CHAPTER 13 SOLUTIONS MP SECTION 13.1 SOLUTIONS...

This preview shows pages 1–3. Sign up to view the full content.

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 Weighing-Put Coins 1-7 on left side of scale, 8-14 on right side of scale, and coins 15-21 off scale. Possible Results of First Weighing Left Side Heavier-Heavy coin is one of coins 1-7 Right Side Heavier-Heavy Coin is one of coins 8-14 Scales Balance-Heavy coin is one of coins 15-21 Second Weighing-Let's suppose we have found that the heavy coin is one of coins 1-7. Then put coins 1-3 on left side of scale, coins 4-6 on right 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
side of the scale, and coin 7 off scale.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 22

CH13 - 1 MP CHAPTER 13 SOLUTIONS MP SECTION 13.1 SOLUTIONS...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online