2019 AMC 10B Problem 6

Both factorials on the left contain $n!$ as a factor:
$$(n+1)! = (n+1)\,n!, \qquad (n+2)! = (n+2)(n+1)\,n!.$$
So the left side is
$$n!\,(n+1)\big[1 + (n+2)\big] = n!\,(n+1)(n+3).$$
Dividing the equation by $n!$ leaves $(n+1)(n+3) = 440$.

We need two numbers that differ by $2$ and multiply to $440$. Since $\sqrt{440} \approx 21$, try $20 \cdot 22 = 440$. Hence $n + 1 = 20$ and $n = 19$.

The sum of the digits of $19$ is $1 + 9 = 10$. The answer is $\boxed{\textbf{(C)}\ 10}$.

Factorials of nearby integers are all multiples of the smallest one, so pulling out $n!$ converts a factorial equation into a polynomial one. A product of two numbers with a fixed difference is found fastest by estimating the square root of the product. Do not forget the final step: the question asks for the digit sum, not $n$.

Expanding gives $n^2 + 4n + 3 = 440$, so $n^2 + 4n - 437 = 0$ and $n = \frac{-4 + \sqrt{16 + 1748}}{2} = \frac{-4 + 42}{2} = 19$.

• Expanding (n+2)! as (n+2) times n! instead of (n+2)(n+1) times n!, which leads to a wrong quadratic.
• Answering $19$ (or a choice near it) instead of the digit sum $10$.