# Difference between revisions of "2005 AIME I Problems/Problem 3"

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== Solution == | == Solution == | ||

− | + | <math>n=p\cdot q</math> or <math>n=p^3</math> 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 <math> {15 \choose 2} =105</math> 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 <math>105+4=109</math> integers that meet the given conditions. | |

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== See also == | == See also == |

## Revision as of 21:53, 28 November 2006

## Problem

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

## Solution

or 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 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 integers that meet the given conditions.