# 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

### Solution 1

Because the sum of $n$ positive integers is $(n)(n+1)/2$, 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 $996 \implies \boxed{\textbf{(B)}}.$

-Lcz

### Solution 2

Following from the fact that $n+1$ must be prime, we can use to answer choices as possible solutions for $n$. $A$, $C$, 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 $996 \implies \boxed{\textbf{(B)}}$.

## See Also

 2019 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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