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.

I will consider the example of $m=3$ to give a sense of why $m$ will work so long as $m-1$ divides 2010. We can write $2187 = 3^7 = \sum_{i=1}^{3^7} 3^0$ . Exchanging three $3^0$ terms for a $3^1$ term leaves the value on the right the same and decreases the number of terms by 2. Thus, we can write $3^7=2187$ using $2187$ terms, or $2185, 2183, \dots$ , where the pattern is that the number of possible terms is $\equiv 1 \pmod{2}$ . Since $2 | 2010 = 2011-1$ , $m=3$ is a value of $m$ for which we can obtain the desired sum. If we run out of $3^0$ terms, we can start exchanging three $3^1$ terms for a $3^2$ term. In general, this exchange will take $m$ terms of $m^k$ and replace them with one $m^{k+1}$ term, thus reducing the number of terms by $(m-1)$ .

2011 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2011 AIME I Problems/Problem 7

Problem 7

Solution

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