A positive integer n satisfies the equation (n+1)! + (n+2)! = n! × 440. What is the sum of the digits of n?

2019 AMC 10B Problem 6

Attempt Problem 6 of the 2019 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

A positive integer

n

satisfies the equation

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

What is the sum of the digits of

n?

2

5

10

12

15

Factor

(n+1)!

and

(n+2)!

in terms of

n!

Divide by

n!

and solve the resulting square equation

We can rewrite the left side as

(n+1)n!+(n+2)(n+1)n!

=((n+2)^2-1)n!,

((n+2)^2-1) n! = 440 n!

Therefore,

(n+2)^2=441,

n=19.

The sum of its digits is

10.

Thus, the answer is C .