2019 AIME II Problems/Problem 9
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, where are primes. No number that is a multiple of can be expressed in the first form because 20 has two primes in its prime factorization, while the first form has only one, and the only integer divisible by that has the second form is , which is 2560, 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, .
Rephrased for clarity by Afly
Solution 3
The divisors of are . must be because . This means that can be exactly or . Because greater exponents of 2 (like ) gives numbers greater than 2019, so we just have the cases below.
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.