PUZZLE CORNER
Kiwanis
Monthly Challenge
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A new puzzle will appear each month.
When you have solved the puzzle, you may submit your solution.
The solution will be given the following month, along with credit to those who have solved the puzzle. Puzzle #3 for May, 2008 “The Politician Puzzle”:
A recent convention had 100 politicians. Each politician was either crooked or honest. We are given the following two additional facts: At least one of the politicians was honest. Given any two of the politicians, at least one of the two was crooked. Can it be determined from these two additional facts how many of the politicians were honest and how many of them were crooked? Text Solution #3: A fairly common answer is "50 honest and 50 crooked." Another rather frequent one is "51 honest and 49 crooked." Both answers are wrong. Now let us see how to find the correct solution. We are given the information that at least one person is honest. Let us pick out anyone honest person, whose name, say, is Frank. Now pick any of the remaining 99; call him John. By the second given condition, at least one of the two men Frank or John - is crooked. Since Frank is not crooked, it must be John. Since John arbitrarily represents any of the remaining 99 men, then each of those 99 men must be crooked. So the answer is that one is honest and 99 are crooked. Another way of proving it is this: the statement that given any two, at least one is crooked, says nothing more or less than that given any two, they are not both honest; in other words, no two are honest. This means that at most one is honest. Also (by the first condition), at least one is honest. Hence exactly one is honest. |
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