2005 AIME I Problems/Problem 3

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Problem

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

Solution

Suppose $n$ is such an integer. Then $n=p\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$.

In the first case, the three proper divisors of $n$ are 1, $p$ and $q$. Thus, we need to pick two prime numbers less than 50. There are fifteen of these (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47) so there are ${15 \choose 2} =105$ numbers of the first type.

In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$. Thus we need to pick a prime number whose square is less than 50. There are four of these (2, 3, 5 and 7) and so four numbers of the second type.

Thus there are $105+4=109$ integers that meet the given conditions.

See also

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