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

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== Solution ==
 
== Solution ==
{{solution}}
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<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.
 
 
The following solution contains errors; please correct it:
 
Having three proper divisors means that there are 4 regular divisors. So the number can be written as <math>\displaystyle p_{1}p_{2}</math> where <math>\displaystyle p_{1}</math> and <math>\displaystyle 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> {15 \choose 2} =105</math> such numbers.
 
  
 
== 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

$n=p\cdot q$ 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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