Difference between revisions of "2019 AIME II Problems/Problem 9"
Line 2: | Line 2: | ||
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. | ||
Latest 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.