2019 AMC 12B Problems/Problem 3
Contents
Problem
If (n+1)!+(n+2)! = n!*440, what is the sum of the digits of n?
Solution
n=19 sum is 10 (SuperWill)
Solution 2
Divide both sides by 
:
factor out 
:
prime factorization of 
and a bit of experimentation gives us 
and 
, so 
.
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
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| All AMC 12 Problems and Solutions | |
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