# Difference between revisions of "2019 AMC 10A Problems/Problem 9"

## Problem

What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?

$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$

### Solution 1

Because the sum of $n$ positive integers is $\frac{(n)(n+1)}{2}$, and we want this to not be a divisor of the $n!$, $n+1$ must be prime. The greatest three-digit integer that is prime is $997$. Subtract $1$ to get $\boxed{\textbf{(B) } 996}$.

### Solution 2

We can use the fact that $n+1$ must be prime to eliminate answer choices as possible values of $n$. $A$, $C$, and $E$ don't work because $n+1$ is even, and $D$ does not work since $999$ is divisible by $9$. Thus, the only correct answer is $\boxed{\textbf{(B) } 996}$.