Difference between revisions of "2019 AIME II Problems/Problem 9"
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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>. | 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 1== |
Every 20-pretty integer can be written in form <math>n = 2^a 5^b k</math>, where <math>a \ge 2</math>, <math>b \ge 1</math>, <math>\gcd(k,10) = 1</math>, and <math>d(n) = 20</math>, where <math>d(n)</math> is the number of divisors of <math>n</math>. Thus, we have <math>20 = (a+1)(b+1)d(k)</math>, using the fact that the divisor function is multiplicative. As <math>(a+1)(b+1)</math> must be a divisor of 20, there are not many cases to check. | Every 20-pretty integer can be written in form <math>n = 2^a 5^b k</math>, where <math>a \ge 2</math>, <math>b \ge 1</math>, <math>\gcd(k,10) = 1</math>, and <math>d(n) = 20</math>, where <math>d(n)</math> is the number of divisors of <math>n</math>. Thus, we have <math>20 = (a+1)(b+1)d(k)</math>, using the fact that the divisor function is multiplicative. As <math>(a+1)(b+1)</math> must be a divisor of 20, there are not many cases to check. | ||
Revision as of 02:30, 16 February 2021
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) | ||
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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