Difference between revisions of "2019 AIME I Problems/Problem 9"
Icematrix2 (talk | contribs) |
Icematrix2 (talk | contribs) |
||
Line 26: | Line 26: | ||
==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}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:38, 16 October 2020
Problem 9
Let denote the number of positive integer divisors of (including and ). Find the sum of the six least positive integers that are solutions to .
Solution
In order to obtain a sum of , we must have:
- either a number with divisors (a fourth power of a prime) and a number with divisors (a prime), or
- a number with divisors (a semiprime or a cube of a prime) and a number with divisors (a square of a prime). (No integer greater than can have fewer than 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 divisors, or a fourth power like with divisors. We then find the smallest such values by hand.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . and both work.
- has two possibilities: and or and . Only works.
- has two possibilities: and or and . Only works.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . Only works.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . Only works.
Having computed the working possibilities, we take the sum of the corresponding values of : . ~Kepy.
Possible improvement: since all primes are odd, their fourth powers are odd as well, which cannot be adjacent to any primes because both of the adjacent numbers will be even. Thus, we only need to check for the fourth power case. - mathleticguyyy
Note: Bashing would work for this problem, but it would be very tedious.
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.