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==Problem 9== | ==Problem 9== | ||
− | Let <math>\tau (n)</math> denote the number of positive integer divisors of <math>n</math>. Find the sum of the six least positive integers <math>n</math> that are solutions to <math>\tau (n) + \tau (n+1) = 7</math>. | + | Let <math>\tau (n)</math> denote the number of positive integer divisors of <math>n</math> (including <math>1</math> and <math>n</math>). Find the sum of the six least positive integers <math>n</math> that are solutions to <math>\tau (n) + \tau (n+1) = 7</math>. |
==Solution== | ==Solution== |
Revision as of 14:42, 14 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 |
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