Difference between revisions of "2019 AMC 12B Problems/Problem 4"

Revision as of 12:27, 14 February 2019

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

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

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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 ==Problem==
+A positive integer <math>n</math> satisfies the equation <math>(n+1)!+(n+2)!=440\cdot n!</math>. What is the sum of the digits of <math>n</math>?
+<math>\textbf{(A) } 2 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 10\qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math>
 ==Solution==