2005 AIME I Problems/Problem 3

Problem

How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50?

Solution

Suppose $n$ is such an integer. Then one of its factors is $1$ , so $n$ must be in the form $n=p\cdot q$ 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 $105+4=\boxed{105}$ integers that meet the given conditions.

2005 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2005 AIME I Problems/Problem 3

Problem

Solution

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