During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

# Difference between revisions of "2017 AIME I Problems/Problem 9"

## Problem 9

Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$.

## Solution 1

Writing out the recursive statement for $a_n, a_{n-1}, \dots, a_{10}$ and summing them gives $$a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10$$ Which simplifies to $$a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)$$ Therefore, $a_n$ is divisible by 99 if and only if $\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$, we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$, we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\boxed{045}$.

## Solution 2

$$a_n \equiv a_{n-1} + n \pmod {99}$$ By looking at the first few terms, we can see that $$a_n \equiv 10+11+12+ \dots + n \pmod {99}$$ This implies $$a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}$$ Since $a_n \equiv 0 \pmod {99}$, we can rewrite the equivalence, and simplify $$0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}$$ $$0 \equiv n(n+1) - 90 \pmod {99}$$ $$0 \equiv n^2+n+9 \pmod {99}$$ $$0 \equiv 4n^2+4n+36 \pmod {99}$$ $$0 \equiv (2n+1)^2+35 \pmod {99}$$ $$64 \equiv (2n+1)^2 \pmod {99}$$ The only squares that are congruent to $64 \pmod {99}$ are $(\pm 8)^2$ and $(\pm 19)^2$, so $$2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}$$ $2n+1 \equiv -8 \pmod {99}$ yields $n=45$ as the smallest integer solution.

$2n+1 \equiv 8 \pmod {99}$ yields $n=53$ as the smallest integer solution.

$2n+1 \equiv -19 \pmod {99}$ yields $n=89$ as the smallest integer solution.

$2n+1 \equiv 19 \pmod {99}$ yields $n=9$ as the smallest integer solution. However, $n$ must be greater than $10$.

The smallest positive integer solution greater than $10$ is $n=\boxed{045}$.