2019 AIME II Problems/Problem 9
Contents
[hide]Problem
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 1
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
Solution 2
For to have exactly
positive divisors,
can only take on certain prime factorization forms: namely,
. No number that is a multiple of
can be expressed in the first form, and the only integer divisible by
that has the second form is
, which is greater than
.
For the third form, the only -pretty numbers are
and
, and only
is small enough.
For the fourth form, any number of the form where
is a prime other than
or
will satisfy the
-pretty requirement. Since
,
. Therefore,
can take on
or
.
Thus, .
Solution 3
The divisors of are
.
must be
because
. This means that
can be exactly
or
.
1. . Then
. The smallest is
which is
. Hence there are no solution in this case.
2. . Then
.
The
case gives one solution,
.
The
case gives
.Using any prime greater than
will make
greater than
.
The answer is .
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 10 | |
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