2011 AIME I Problems/Problem 7

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

First I show that $m-1$ must divide $2010$ . Consider the desired equation $\pmod{m-1}$ . The left side is $\equiv 1$ , whereas the right side is $\equiv 2011$ . Thus, we have $1 \equiv 2011 \pmod{m-1}$ , so $m-1$ must divide 2010.

Now let us prove that any m where m-1 divides 2010 works. Suppose we have such an m. Consider the equation m^k = m^k, for sufficiently large k. Replace m^k on the right hand side by m^(k-1)+m^(k-1)+...m^(k-1), with k terms. We have increased the number of terms by k-1. By repeating this process, we can generate an expression on the rhs with 1+ A(k-1) terms, where A is any positive integer. Since k-1 divides 2010, we can choose A to make the value of 1+A(k-1) equal to 2011.

2011 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2011 AIME I Problems/Problem 7

Problem 7

Solution

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