2019 AMC 12B Problem 4 — Solution | LIVE by Po-Shen Loh

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2019 AMC 12B Problem 4

Attempt Problem 4 of the 2019 AMC 12B 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 12B solutions, or check the answer key.

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

4.

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

$2$

$5$

$10$

$12$

$15$

Answer: C

Concepts:factorial quadratic

Difficulty rating: 1200

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

So $n^2+4n-437=0,$ which factors as $(n-19)(n+23)=0,$ giving $n=19.$ Its digit sum is $1+9=10.$

Thus, C is the correct answer.

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