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?

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

Solutions

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 $996 \implies \boxed{\textbf{(B)}}.$

-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 $996 \implies \boxed{\textbf{(B)}}$ .

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2019 AMC 10A Problems/Problem 9

Contents

Problem

Solutions

Solution 1

Solution 2

See Also