2005 AIME I Problems/Problem 3

Revision as of 19:58, 28 October 2006 by Bpms (talk | contribs) (A solution to an AIME problem.)

Problem

How many positive integers have exactly three proper divisors, each of which is less than 50?

Solution

Having three proper divisors means that there are 4 regular divisors. So the number can be written as $<math>p_{1}p_{2}$</math> where $<math>p_{1}$</math> and $<math>p_{2}$</math> are primes. The primes under fifty are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. There are 15 of them. So there are $<math>\binom{15}{2}=105$</math> numbers.

See also