# Splitting Piles

Select a spectator from the audience and present him (or her) with a pile of chips, counters, pebbles or any other kind of small objects. Explain to the person and the audience what you expect them to do. Leave the stage while they are following your bid. Ask them to call you when finished. As the result of their activity, they will produce a number. Upon your stepping back on the stage you declare that you know what the number has been generated, and eventually reveal the number.

This is what the fellow has to do in your absence. He (or she) will be splitting a pile at a time into two piles, multiply the numbers of chips in the two new piles and keep adding the results. The process stops when there is no pile with more than 1 chip.

 Piles Which is broken What's added Total 9 9 3*6 18 3,6 3 1*2 20 1,2,6 6 3*3 29 1,2,3,3 3 1*2 31 1,2,1,2,3 2 1*1 32 1,1,1,1,2,3 2 1*1 33 1,1,1,1,1,1,3 3 1*2 35 1,1,1,1,1,1,1,2 2 1*1 36 1,1,1,1,1,1,1,1,1 - - -

When you return you will somehow know that the result is 36! The secret is that the result is always the same: it does not depend on how the piles are split; but only on the initial size of the very first pile. Obviously, if you want to keep your audience in the dark, do not perform the trick with the same starting number of chips more than once.There is a very simple proof of this fact.

Here is the proof that uses mathematical induction.

Assume we start with N = 2 chips. The only way to split such a pile is to halve it into two piles of 1 chip each. The computed number is just 1. Of course it's independent of how you split the pile; for there is just one way to perform this feat. Note that starting with N = 1 leads to the number 0. We split nothing. One can argue it's the same as having a pile with 0 chips which contributes 0 to the total regardless of the number of chips in other piles.

Now, assume that the result has been established for all numbers less than N and let there be N > 2 chips in the original pile. Split it into two with n and m chips, respectively. We have n, m > 0 and n + m = N. By the inductive assumption, proceeding with the first pile we'll get the number n(n - 1)/2 regardless of how we actually proceed. For the second pile, we'll get m(m - 1)/2. The total is mn + n(n - 1)/2 + m(m - 1)/2 which, after a series of simplifications, yields (m + n)(m + n - 1)/2 = N(N - 1)/2 which

1. is dependent on neither original nor consecutive splits
2. is exactly the number we expected.

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