Problem:
By a proper divisor of a natural number we mean a positive integral divisor other than and the number itself. A natural number greater than will be called "nice" if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?
Solution:
Let be a positive integer, and let be its divisors in ascending order. Then . For to be nice, we must have . Moreover, must be prime, for otherwise, the proper divisors of would have appeared in the listing above between and . Similarly, is either a prime or the square of , for otherwise could not be the only divisor between and . Therefore, is either the product of two distinct primes or -- being the product of a prime and its square -- is the cube of a prime.
In view of the above, one can easily list the first ten nice numbers. They are: and . Their sum is .
The problems on this page are the property of the MAA's American Mathematics Competitions