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  1. 1 point
    You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes). Using a balance scale twice, find seven heavy dimes. EDIT: for clarification
  2. 1 point
  3. 1 point
    Ten years ago I called attention to a number that when divided by a single integer p it left a remainder of p-1. (Help, a remainder is chasing me) Here is a chance to construct a nine-digit number, a permutation of { 1 2 3 4 5 6 7 8 9 } that has no remainders, sort of. The task is to permute { 1 2 3 4 5 6 7 8 9 } to create a number whose first n digits is a multiple of n for any single-digit n. For example, consider 123654987. Its first 2 digits (12) are divisible by 2. It's first 5 digits (12365) are divisible by 5. However this is not a solution, since 1236549 is not a multiple of 7.
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