Problem:
A positive integer n satisfies the equation (n+1)!+(n+2)!=440â‹…n!. What is the sum of the digits of n?
Answer Choices:
A. 2
B. 5
C. 10
D. 12
E. 15
Solution:
Dividing the given equation by n!, simplifying, and completing the square yields
(n+1)+(n+2)(n+1)n2+4n+3n2+4n+4(n+2)2​=440=440=441=212​
Thus n+2=21 and n=19. The requested sum of the digits of n is 1+9=(C)10​.
The problems on this page are the property of the MAA's American Mathematics Competitions