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LostInParadise's avatar

Is this a good math problem?

Asked by LostInParadise (26888points) 3 days ago

I just thought of it. It does not require any advanced math, just some simple reasoning.

You have four coins a, b, c and d. The weights are related by a < b < c < d, though you don’t know the actual weights. You remember that one of the coins weighs exactly twice as much as one of the other coins, but you forgot which two coins they are.

You take a guess and place two a coins in one pan of a balance scale and one c coin in the other. The side with the two a coins goes down, meaning that c weighs less than twice as much as a. How can you determine which coin weighs twice as much as which other coin using a single additional weighing?

Hint (if you need it): After the first weighing, which other pairs of coins are there where you know the heavier of the two coins is less than twice as heavy as the lighter of the two coins?

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1 Answer

LostInParadise's avatar

Well I thought this was a good way to distract oneself from corona.

Here is the answer, for anyone interested. If c is less than twice a then c is less than twice b, since b is greater than a, and b is less than twice a, since b is less than c. That eliminates all cases where b or c is the heavier coin, and of course there are no cases where a is the heavier coin. That means all the remaining cases are where d is the heavier coin, and a, b or c is the lighter coin.

Weight d against two b coins. If it balances then d is twice b. If d is less than twice b then d will also be less than twice c, so the two coins would be d and a. If d is more than twice b then it will also be more than twice a, so the two coins would be d and c.

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