2019 AMC 10A Problem 9

The sum is $\frac{n(n+1)}{2}$ and the product is $n!$. Their quotient is
$$\frac{n!}{n(n+1)/2} = \frac{2\,(n-1)!}{n+1},$$
so the sum divides the product exactly when $n + 1$ divides $2(n-1)!$.

If $n + 1$ is a prime $p > 2$, then $p$ divides none of the factors $1, 2, \dots, n - 1$ nor the $2$, so $p$ does not divide $2(n-1)!$: the sum is not a divisor. If $n + 1$ is composite, its prime factors are all at most $(n+1)/2 \le n - 1$ and appear (with room to spare) inside $2(n-1)!$, so the sum does divide the product.

So we need the largest three-digit $n$ with $n + 1$ prime. Check downward: $1000$, $999 = 27 \cdot 37$, and $998 = 2 \cdot 499$ are composite, while $997$ is prime (it has no prime factor up to $31$). Hence $n = 996$.

The answer is $\boxed{\textbf{(B)}\ 996}$.

Dividing the factorial by the triangular number leaves $2(n-1)!/(n+1)$, which turns the question into "does $n + 1$ divide a product of smaller numbers?" That is where primes stand out: a prime only divides a product if it divides one of the factors. Reducing a fraction before reasoning about divisibility is the habit to take away.

Work directly from the choices: for each $n$ from $999$ down, ask whether $n + 1$ has all its prime factors below $n$. $1000 = 2^3 5^3$, $999 = 3^3 \cdot 37$ and $998 = 2 \cdot 499$ all do (and the powers are easily covered), so those $n$ divide; $997$ is prime, so $n = 996$ is the first failure.

• Testing n = 999 and 998 too casually: 1000 and 999 are composite, so they do divide, and the first failure is n + 1 = 997.
• Choosing $997$ because it is the prime that appears, forgetting that the condition concerns $n + 1$, not $n$.