Difference between revisions of "2019 AMC 12B Problems/Problem 3"
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==Solution 2== | ==Solution 2== | ||
| − | Divide both sides by n!: | + | Divide both sides by <math>n!</math>: |
| − | (n+1)+(n+1)(n+2)=440 | + | <math>(n+1)+(n+1)(n+2)=440</math> |
| − | factor out (n+1): | + | factor out <math>(n+1)</math>: |
| − | (n+1)*(n+3)=440 | + | <math>(n+1)*(n+3)=440</math> |
| − | prime factorization of 440 and a bit of experimentation gives us n+1=20 and n+3=22, so \boxed{n=19}. | + | prime factorization of <math>440</math> and a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>\boxed{n=19}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}} | {{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:34, 14 February 2019
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 |
| 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 | |
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