2019 AIME I Problems/Problem 9
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
In order to obtain a sum of 7, we must have:
- either a number with 5 divisors (a fourth power of a prime) and a number with 2 divisors (a prime), or
- a number with 4 divisors (a semiprime) and a number with 3 divisors (a square of a prime). (No integer greater than 1 can have fewer than 2 divisors.)
Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square like with 3 divisors, or a fourth power like with 5 divisors. We then find the smallest such values by hand.
- has two possibilities: 3 and 4, or 4 and 5. Neither works.
- has two possibilities: 8 and 9, or 9 and 10. (8,9) and (9,10) both work.
- has two possibilities: 15 and 16, or 16 and 17. Only (16,17) works.
- has two possibilities: 24 and 25, or 25 and 26. Only (25,26) works.
- has two possibilities: 48 and 49, or 49 and 50. Neither works.
- has two possibilities: 80 and 81, or 81 and 82. Neither works.
- has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works.
- has two possibilities: 168 and 169, or 169 and 170. Neither works.
- has two possibilities: 288 and 289, or 289 and 290. Neither works.
- has two possibilities: 360 and 361, or 361 and 362. Only (361,362) works.
Having computed the working possibilities, we take the sum of the corresponding values of : . ~Kepy.
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
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Followed by Problem 10 | |
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