2019 AMC 10B Problem 6

Factorials of nearby numbers share a lot of factors. Seeing that first turns this into an ordinary equation.

1. Write each factorial in terms of $n!$:
$$(n+1)! = (n+1) \cdot n!, \qquad (n+2)! = (n+2)(n+1) \cdot n!.$$
For example, with $n = 3$: $4! = 4 \cdot 3!$ and $5! = 5 \cdot 4 \cdot 3!$.
2. So the left side is
$$(n+1)\,n! + (n+2)(n+1)\,n! = n!\,(n+1)\big(1 + (n+2)\big) = n!\,(n+1)(n+3).$$
3. The equation becomes $n!\,(n+1)(n+3) = n! \cdot 440$. Divide both sides by $n!$:
$$(n+1)(n+3) = 440.$$
4. We want two numbers that differ by $2$ and multiply to $440$. They should be close to $\sqrt{440}$, which is a bit less than $21$. Try $20 \cdot 22 = 440$. It works.
5. So $n + 1 = 20$ and $n = 19$.
6. The question asks for the sum of the digits of $n$: $1 + 9 = 10$.

The answer is $\boxed{\textbf{(C)}\ 10}$.

A bigger factorial is always a smaller factorial times some extra numbers, so pulling out the common $n!$ removes the factorials completely. Then a product of two numbers that differ by $2$ is found by guessing near the square root. Watch the last line of the question: it wants the digit sum, not $n$, and $19$ is not among the choices anyway.

Multiply out instead of guessing: $(n+1)(n+3) = n^2 + 4n + 3 = 440$, so $n^2 + 4n - 437 = 0$. The quadratic formula gives
$$n = \frac{-4 + \sqrt{16 + 1748}}{2} = \frac{-4 + 42}{2} = 19.$$