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

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==Problem==
 
==Problem==
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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>?
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<math>\textbf{(A) } 2 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 10\qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math>
  
 
==Solution==
 
==Solution==

Revision as of 12:27, 14 February 2019

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

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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