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2005 AIME I Problems/Problem 3

Revision as of 21:53, 28 November 2006 by 1337h4x (talk | contribs)

Problem

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

Solution

$n=p\cdot q$ or $n=p^3$ where p and q are distinct primes. In the first case, the three proper divisors of n are 1, p, and q. Because there are 15 prime numbers less than 50, there are ${15 \choose 2} =105$ numbers of the first type.There are four integers of the second type because 2, 3, 5, and 7 are the only primes with squares less than 50. Thus there are $105+4=109$ integers that meet the given conditions.

See also

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