Difference between revisions of "2005 AIME I Problems/Problem 3"
(A solution to an AIME problem.) |
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== Solution == | == Solution == | ||
| + | Having three proper divisors means that there are 4 regular divisors. So the number can be written as <math><math>p_{1}p_{2}</math></math> where <math><math>p_{1}</math></math> and <math><math>p_{2}</math></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><math>\binom{15}{2}=105</math></math> numbers. | ||
== See also == | == See also == | ||
* [[2005 AIME I Problems]] | * [[2005 AIME I Problems]] | ||
Revision as of 18:58, 28 October 2006
Problem
How many positive integers have exactly three proper divisors, each of which is less than 50?
Solution
Having three proper divisors means that there are 4 regular divisors. So the number can be written as 
</math> where 
</math> and 
</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> numbers.
