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You are blindfolded before a table. On the table are a very large number of pennies. You are told that 2007 of the pennies are heads up and the rest are tails up. How can you divide the entire set of pennies into two subgroups, each with the same number of heads facing up?

2007-12-18 05:08:39 · 2 answers · asked by BadCompany 1 in Science & Mathematics Mathematics

2 answers

Take any 2007 pennies. Of these, X wil be heads up and 2007 - X will be tails up. The remaining pennies have 2007 - X that are heads up. In your group, flip every coin. Now your group has 2007 - X heads up, which matches the number of heads-up coins in the leftover group.

2007-12-18 05:12:30 · answer #1 · answered by jgoulden 7 · 2 0

The logic from the previous answer is very clever.

2007-12-18 05:48:43 · answer #2 · answered by matt p 1 · 0 0

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