2019 AMC 12B Problems/Problem 4

Problem

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

$\textbf{(A) } 2 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 10\qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$

Solution

Dividing both sides by $n!$ gives $(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23).$ Since $n$ is positive, $n=19$ . The answer is $1+9=10\Rightarrow \boxed{C}.$

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 12 Problems and Solutions

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2019 AMC 12B Problems/Problem 4

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