# Difference between revisions of "2018 AMC 10B Problems/Problem 21"

## Problem

Mary chose an even $4$-digit number $n$. She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,...,\dfrac{n}{2},n$. At some moment Mary wrote $323$ as a divisor of $n$. What is the smallest possible value of the next divisor written to the right of $323$?

$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$

## Solution 1

Prime factorizing $323$ gives you $17 \cdot 19$. The desired answer needs to be a multiple of $17$ or $19$, because if it is not a multiple of $17$ or $19$, the LCM, or the least possible value for $n$, will not be more than 4 digits. Looking at the answer choices, $\fbox{(C) 340}$ is the smallest number divisible by $17$ or $19$. Checking, we can see that $n$ would be $6460$.

## Solution 2

Let the next largest divisor be $k$. Suppose $\gcd(k,323)=1$. Then, as $323|n, k|n$, therefore, $323\cdot k|n.$ However, because $k>323$, $323k>323\cdot 324>9999$. Therefore, $\gcd(k,323)>1$. Note that $323=17\cdot 19$. Therefore, the smallest the gcd can be is $17$ and our answer is $323+17=\boxed{\text{(C) }340}$.