Difference between revisions of "2019 AIME I Problems/Problem 9"
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==Solution== | ==Solution== | ||
+ | Essentially, you realize that to get 7 you need an odd amount of divisors + and even amount of divisors. This means that one of our n needs to be a square. Furthermore it must either be a prime squared to get 3 divisors or a prime to the fourth to get 5 divisors. Any more factors in a square would be to large. Thus n/n+1 is in the form p^2 or p^4. The rest of the solution is bashing left to the reader. | ||
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+ | ~~ paliwalar.21 | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=8|num-a=10}} | {{AIME box|year=2019|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:28, 14 March 2019
The 2019 AIME I takes place on March 13, 2019.
Problem 9
Let denote the number of positive integer divisors of . Find the sum of the six least positive integers that are solutions to .
Solution
Essentially, you realize that to get 7 you need an odd amount of divisors + and even amount of divisors. This means that one of our n needs to be a square. Furthermore it must either be a prime squared to get 3 divisors or a prime to the fourth to get 5 divisors. Any more factors in a square would be to large. Thus n/n+1 is in the form p^2 or p^4. The rest of the solution is bashing left to the reader.
~~ paliwalar.21
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.