## Solution to Puzzle #91: Pebble Piles

I again got a very high response this week – thanks again!

There was only one correct answer from Gyora – well done!

I am taking the liberty of reproducing the answer from Gurmeet, whose site is the source of this puzzle.

Since all three piles are odd, the first operation must be merger of two existing piles, resulting in an even-sized pile and an odd-sized pile. Let g denote the greatest common divisor of these two pile sizes. Since one pile is odd sized and the other pile is even sized, g must be odd. For odd g, no matter what sequence of operations is carried out (merger of two existing piles or division of an even pile into two equal piles), every pile size will continue to be a multiple of g. If g > 1, then we can never reach a configuration where all piles are size 1. Now, starting with {5, 49, 51}, three states are possible: {54, 51}, {5, 100} and {49, 56}. The GCD is 3, 5 and 7, respectively, in these three states. So we can never get 105 piles with 1 pebble each.

Hope you enjoyed the puzzle!

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