2019 AMC 12B Problem 4

A positive integer

n

satisfies the equation

(n+1)!+(n+2)!=440\cdot n!.

What is the sum of the digits of

n?

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Solution

Factor the left side:

\begin{gathered} (n+1)!+(n+2)! \\ =(n+1)!\,[1+(n+2)] \\ =(n+1)!\,(n+3). \end{gathered}

Dividing both sides by

n!

and using

(n+1)!=(n+1)\,n!

gives

(n+1)(n+3)=440.

n^2+4n-437=0,

which factors as

(n-19)(n+23)=0,

giving

n=19.

Its digit sum is

1+9=10.

Thus, C is the correct answer.