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

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== Solution ==
 
== Solution ==
 
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The following solution contains errors; please correct it:
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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.
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== See also ==
 
== See also ==
 
* [[2005 AIME I Problems/Problem 2 | Previous problem]]
 
* [[2005 AIME I Problems/Problem 2 | Previous problem]]

Revision as of 17:07, 29 October 2006

Problem

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

Solution

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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 $\displaystyle p_{1}p_{2}$ where $\displaystyle p_{1}$ and $\displaystyle p_{2}$ 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 ${15 \choose 2} =105$ such numbers.

See also

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