# 2018 AMC 10B Problems/Problem 21

The following problem is from both the 2018 AMC 12B #19 and 2018 AMC 10B #21, so both problems redirect to this page.

## Problem

Mary chose an even $4$-digit number $n$. She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$. 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 (Inequalities)

Let $d$ be the next divisor written to the right of $323.$

If $\gcd(323,d)=1,$ then $$n\geq323d>323^2>100^2=10000,$$ which contradicts the precondition that $n$ is a $4$-digit number.

It follows that $\gcd(323,d)>1.$ Since $323=17\cdot19,$ the smallest possible value of $d$ is $17\cdot20=\boxed{\textbf{(C) } 340},$ from which $$n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.$$ ~MRENTHUSIASM ~tdeng

## Solution 2 (Inequalities)

Let $d$ be the next divisor written to the right of $323.$

Since $n$ is even and $323=17\cdot19,$ we have $n=2\cdot17\cdot19\cdot k=646k$ for some positive integer $k.$ Moreover, since $1000\leq n\leq9998,$ we get $2\leq k\leq15.$ As $d>323,$ it is clear that $d$ must be divisible by $17$ or $19$ or both.

Therefore, the smallest possible value of $d$ is $17\cdot20=\boxed{\textbf{(C) } 340},$ from which $$n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.$$ ~MRENTHUSIASM ~bjhhar

## Solution 3 (Quick)

The prime factorization of $323$ is $17 \cdot 19$. Our answer must be a multiple of either $17$ or $19$ or both. Since $17 < 19$, the next smallest divisor that is divisble by $17$ would be $323 + 17 = \boxed{\textbf{(C) } 340}$

Since prime factorizing $323$ gives you $17 \cdot 19$, the desired answer needs to be a multiple of $17$ or $19$, this is because if it is not a multiple of $17$ or $19$, $n$ will be more than a $4$ digit number. For example, if the answer were to instead be $324$, $n$ would have to be a multiple of $2^2\cdot3^4\cdot17\cdot19$ for both $323$ and $324$ to be a valid factor, meaning $n$ would have to be at least $104652$, which is too big. Looking at the answer choices, $\textbf{(A)}$ and $\textbf{(B)}$ are both not a multiple of neither $17$ nor $19$, $\textbf{(C)}$ is divisible by $17$. $\textbf{(D)}$ is divisible by $19$, and $\textbf{(E)}$ is divisible by both $17$ and $19$. Since $\boxed{\textbf{(C) } 340}$ is the smallest number divisible by either $17$ or $19$ it is the answer. Checking, we can see that $n$ would be $6460$, a $4$-digit number. Note that $n$ is also divisible by $2$, one of the listed divisors of $n$. (If $n$ was not divisible by $2$, we would need to look for a different divisor.)

-Edited by Mathandski

Note that $323$ multiplied by any of the answer choices results in a $5$ or $6$-digit $n$. So, we need a choice that shares a factor(s) with $323$, such that the factors we'll need to add to the prime factorization of $n$ (in result to adding the chosen divisor) won't cause our number to multiply to more than $4$ digits. The prime factorization of $323$ is $17\cdot19$, and since we know $n$ is even, our answer needs to be

• even
• has a factor of $17$ or $19$

We see $340$ achieves this and is the smallest to do so ($646$ being the other). So, we get $\boxed{\textbf{(C) } 340}$.

~OGBooger (Solution)

~Pearl2008 (Minor Edits)

~bunny1