A problem from the Leningrad Mathematical Olympiad: You have a set of 101 coins, and you know that it contains one counterfeit coin X. The 100 genuine coins all have the same weight, which is different from that of X. Using only two weighings in an equal-arm balance, how can you determine whether X is heavier or lighter than the genuine coins?
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Divide the coins arbitrarily into piles of 30, 30, and 41 coins, calling these respectively A, B, and C. Now weigh A against B. If these balance, then X must be in C; combine A and B and weigh 41 of these coins against the 41 coins in C and you’ll find out whether X is heavy or light.
If A and B don’t balance, then one of those groups contains the bad coin. For definiteness, suppose that A goes down in the weighing. That means either that A contains the bad coin and it’s heavy or that B contains the bad coin and it’s light. So determining where X is will also tell us its relative weight.
Divide A in half and weigh these halves against each other. If they balance then the coins are all good and X must be in B. If they don’t balance then X is in A.
(Obviously the same plan will work with certain pile sizes other than 30, 30, and 41, but that arrangement will do the job.)
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