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

Latest revision as of 14:28, 14 February 2019

Redirect to:

@@ Line 1: / Line 1: @@
-==Problem==
+#REDIRECT[[2019_AMC_10B_Problems/Problem_6]]
-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==
-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)=0.</cmath>
-Since <math>n</math> is positive, <math>n=19</math>. The answer is <math>1+9=10\Rightarrow \boxed{C}.</math>
-==See Also==
-{{AMC12 box|year=2019|ab=B|num-b=3|num-a=5}}
-{{MAA Notice}}