# 1987 AIME Problems/Problem 11

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## Problem

Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers.

## Solutions

### Solution 1

Let us write down one such sum, with $m$ terms and first term $n + 1$: $3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)$.

Thus $m(2n + m + 1) = 2 \cdot 3^{11}$ so $m$ is a divisor of $2\cdot 3^{11}$. However, because $n \geq 0$ we have $m^2 < m(m + 1) \leq 2\cdot 3^{11}$ so $m < \sqrt{2\cdot 3^{11}} < 3^6$. Thus, we are looking for large factors of $2\cdot 3^{11}$ which are less than $3^6$. The largest such factor is clearly $2\cdot 3^5 = 486$; for this value of $m$ we do indeed have the valid expression $3^{11} = 122 + 123 + \ldots + 607$, for which $k=\boxed{486}$.

### Solution 2

First note that if $k$ is odd, and $n$ is the middle term, the sum is equal to kn. If $k$ is even, then we have the sum equal to $kn+k/2$ which is going to be even. Since $3^{11}$ is odd, we see that $k$ is odd.

Thus, we have $nk=3^{11} \implies n=3^{11}/k$. Also, note $n-(k+1)/2=0 \implies n=(k+1)/2.$ Subsituting $n=3^{11}/k$, we have $k^2+k=2*3^{11}$. Proceed as in solution 1.

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