Two-Toned

A problem from the 2004 Moscow Mathematical Olympiad:

An arithmetic progression consists of integers. The sum of its first n terms is a power of two. Prove that n is also a power of two.

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If a and b are the first and nth terms of the progression, then the sum of the first n terms is S = (a + b)n/2. So 2S is divisible by n, and since 2S is a power of 2, so is n.