2019 AMC 10A Problems/Problem 9

Revision as of 16:47, 12 February 2019 by Aiwen (talk | contribs) (Solution 2)

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

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

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 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

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