Difference between revisions of "1995 AIME Problems/Problem 6"
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== Solution == | == Solution == | ||
| + | We know that <math>n^2</math> must have <math>63\times 39</math> factors by its prime factorization. There are <math>\frac{63\times 39-1}{2} = 1228</math> factors of <math>n^2</math> that are less than <math>n</math>, because if they form pairs <math>a</math>, then there is one factor per pair that is less than <math>n</math>. There are <math>32\times20-1 = 639</math> factors of <math>n</math> that are less than <math>n</math> itself. These are also factors of <math>n^2</math>. Therefore, there are <math>1228-639=\fbox{539}</math> factors of <math>n</math> that does not divide <math>n</math>. | ||
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== See also == | == See also == | ||
* [[1995 AIME Problems/Problem 5 | Previous problem]] | * [[1995 AIME Problems/Problem 5 | Previous problem]] | ||
* [[1995 AIME Problems/Problem 7 | Next problem]] | * [[1995 AIME Problems/Problem 7 | Next problem]] | ||
* [[1995 AIME Problems]] | * [[1995 AIME Problems]] | ||
Revision as of 17:58, 8 February 2007
Problem
Let 
How many positive integer divisors of 
are less than 
but do not divide ?
Solution
We know that 
must have 
factors by its prime factorization. There are 
factors of that are less than 
, because if they form pairs 
, then there is one factor per pair that is less than . There are 
factors of that are less than itself. These are also factors of . Therefore, there are 
factors of that does not divide .
