2011 AIME I Problems/Problem 7

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Problem 7

Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$, $x_1$ , $\dots$ , $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\]

Solution

NOTE: This solution is incomplete. Please help make it better.

This formula only works if $m$ is exactly 1 more than a factor of 2010. (Someone else should insert a convincing proof here; I forgot exactly what I did. However, it involved starting with each $x_k$ equal to an unspecified large number $q$, and then decreasing the powers of a certain number of the terms [equal to $m$ to certain powers] by 1 repeatedly in certain ways until the expression becomes $m^p$$m^q$ for some integer p.)

Since 2010 factors as $2^1 3^1 5^1 67^1$, there are $2^4=\boxed{016}$ such factors.

See also

2011 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions