Create a pile of m coins and flip all of them over.
Like I said, I only came to this based on the equation of heads in each pile. It was obvious that the solution would involve flipping them over in some pattern, so I started working out equations for various combinations. Flipping over all the coins in a pile looked promising because that was the point at which the permutations of possible changes to the number of heads all converged back to a single answer. After that, it was easy to see that the relevant term (how many heads happened to be in one of the piles) canceled itself out if one pile was inverted, and then it was just a matter of solving for the correct pile size.
Having said that, it makes real physical sense. If the pile 1 has m coins and has x heads in it, then pile 2 has m - x. If you flip all the coins in pile 1, then it must now also have m - x heads now. QED.
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Like I said, I only came to this based on the equation of heads in each pile. It was obvious that the solution would involve flipping them over in some pattern, so I started working out equations for various combinations. Flipping over all the coins in a pile looked promising because that was the point at which the permutations of possible changes to the number of heads all converged back to a single answer. After that, it was easy to see that the relevant term (how many heads happened to be in one of the piles) canceled itself out if one pile was inverted, and then it was just a matter of solving for the correct pile size.
Having said that, it makes real physical sense. If the pile 1 has m coins and has x heads in it, then pile 2 has m - x. If you flip all the coins in pile 1, then it must now also have m - x heads now. QED.