2008 Mock ARML 2 Problems/Problem 2
Given that the sum of all positive integers with exactly two proper divisors, each of which is less than , is , find the sum of all positive integers with exactly three proper divisors, each of which is less than (a proper divisor of is a positive integer that divides but is not equal to ).
Let be primes and . Then we note that . In other words, only perfect squares of primes have exactly two proper divisors (hence ), and the desired sum consists of all products of two distinct primes and perfect cubes of primes.
The former can be found by using the well-known expansion , or Thus, this sum is .
The perfect cubes whose divisors are all less than are , and so the answer is .
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