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

The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$, and we want this not to be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$, which means they would be able to cancel with the factors in $n!$. Thus, the sum of $n$ positive integers would be a divisor of $n!$ when $n+1$ is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, $n+1$ must instead be prime. The greatest three-digit integer that is prime is $997$, so we subtract $1$ to get $n=\boxed{\textbf{(B) } 996}$.

Solution 2

As in Solution 1, we deduce that $n+1$ must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of $n$. Choices $A$, $C$, and $E$ don't work because $n+1$ is even, and all even numbers are divisible by two, which makes choices $A$, $C$, and $E$ composite and not prime. Choice $D$ also does not work since $999$ is divisible by $9$, which means it's a composite number and not prime. Thus, the correct answer must be $\boxed{\textbf{(B) } 996}$.

Solution 3 (Elimination)

The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ and the product of the positive integers upto $n$ is $n!$. The quotient of the two is -

$\frac{(2)(n!)}{(n)(n+1)}$

which simplifies to $\frac{(2)((n-1)!)}{n+1}$. Thus, $n+1$ must be odd for the remainder to not be 0 (as $2$ will multiply with some number in $n!$, cancelling out $n+1$ if it is even, which leaves us with the answer choices $n = 996$ and $n = 998$. Notice that $n + 1$ must also be prime as otherwise there will be a factor of $n + 1$ in $2$ x $n!$ somewhere. So either $997$ or $999$ must be prime - $999$ is obviously not prime as it is divisible by 9, so our answer should be $n$ where $n + 1 = 997$, and so our answer is $n = 996$ or $\boxed{\textbf{(B) } 996}$.

Video Solutions

~savannahsolver

~ pi_is_3.14

Video Solution 3

Education, the Study of Everything