|
|
| (One intermediate revision by one other user not shown) |
| 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).</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}}
| |
