Difference between revisions of "2019 AMC 12B Problems/Problem 4"
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==Problem== | ==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>? | ||
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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 13:27, 14 February 2019
Problem
A positive integer 
satisfies the equation 
. What is the sum of the digits of ?
Solution
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
