Difference between revisions of "2019 AIME II Problems/Problem 9"
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==Problem 9== | ==Problem 9== | ||
− | Call a positive integer <math>n</math> <math>k</math>-<i>pretty</i> if <math>n</math> has exactly <math>k</math> positive divisors and <math>n</math> is divisible by <math>k</math>. For example, <math>18</math> is <math>6</math>-pretty. Let <math>S</math> be the sum of positive integers less than <math>2019</math> that are <math>20</math>-pretty. Find <math>\ | + | Call a positive integer <math>n</math> <math>k</math>-<i>pretty</i> if <math>n</math> has exactly <math>k</math> positive divisors and <math>n</math> is divisible by <math>k</math>. For example, <math>18</math> is <math>6</math>-pretty. Let <math>S</math> be the sum of positive integers less than <math>2019</math> that are <math>20</math>-pretty. Find <math>\tfrac{S}{20}</math>. |
==Solution== | ==Solution== |
Revision as of 18:25, 22 March 2019
Problem 9
Call a positive integer
-pretty if
has exactly
positive divisors and
is divisible by
. For example,
is
-pretty. Let
be the sum of positive integers less than
that are
-pretty. Find
.
Solution
Every 20-pretty integer can be written in form , where
,
,
, and
, where
is the number of divisors of
. Thus, we have
, using the fact that the divisor function is multiplicative. As
must be a divisor of 20, there are not many cases to check.
If , then
. But this leads to no solutions, as
gives
.
If , then
or
. The first case gives
where
is a prime other than 2 or 5. Thus we have
. The sum of all such
is
. In the second case
and
, and there is one solution
.
If , then
, but this gives
. No other values for
work.
Then we have .
-scrabbler94
See Also
2019 AIME II (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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