2019 AIME I Problems/Problem 9
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 | |
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