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 .
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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 . 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) | ||
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Followed by Problem 10 | |
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