Difference between revisions of "2019 AMC 12B Problems/Problem 3"
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n=19 | n=19 | ||
sum is 10 (SuperWill) | sum is 10 (SuperWill) | ||
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| + | ==Solution 2== | ||
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| + | Divide both sides by n!: | ||
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| + | (n+1)+(n+1)(n+2)=440 | ||
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| + | factor out (n+1): | ||
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| + | (n+1)*(n+3)=440 | ||
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| + | prime factorization of 440 and a bit of experimentation gives us n+1=20 and n+3=22, so \boxed{n=19}. | ||
==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 12:33, 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 n!:
(n+1)+(n+1)(n+2)=440
factor out (n+1):
(n+1)*(n+3)=440
prime factorization of 440 and a bit of experimentation gives us n+1=20 and n+3=22, so \boxed{n=19}.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
