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

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

Revision as of 12:31, 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

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

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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