How many reduced fractions a/b such that ab=20! and 0 < a/b < 1?

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(a, b have to be integers)



Assuming a,b both positive, we get
$ab<b^2$



Therefore, $b> sqrt20!$



Similarly, $a < sqrt20!$



I am stuck after this. Help?




combinatorics factorial fractions




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  • Let $A$ be the set of all divisors $a$ of $20!$ such that $a<sqrt20!$. Define $B$ in a similar way. Now $D=Acup B$ where $D$ is the set of all divisors
    – Lozenges
    Aug 17 at 6:48














up vote
1
down vote

favorite












(a, b have to be integers)



Assuming a,b both positive, we get
$ab<b^2$



Therefore, $b> sqrt20!$



Similarly, $a < sqrt20!$



I am stuck after this. Help?




combinatorics factorial fractions




share|cite|improve this question






















  • Let $A$ be the set of all divisors $a$ of $20!$ such that $a<sqrt20!$. Define $B$ in a similar way. Now $D=Acup B$ where $D$ is the set of all divisors
    – Lozenges
    Aug 17 at 6:48












up vote
1
down vote

favorite









up vote
1
down vote

favorite











(a, b have to be integers)



Assuming a,b both positive, we get
$ab<b^2$



Therefore, $b> sqrt20!$



Similarly, $a < sqrt20!$



I am stuck after this. Help?




combinatorics factorial fractions




share|cite|improve this question














(a, b have to be integers)



Assuming a,b both positive, we get
$ab<b^2$



Therefore, $b> sqrt20!$



Similarly, $a < sqrt20!$



I am stuck after this. Help?





combinatorics factorial fractions





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 17 at 5:04

























asked Aug 17 at 5:01









redx

314




314











  • Let $A$ be the set of all divisors $a$ of $20!$ such that $a<sqrt20!$. Define $B$ in a similar way. Now $D=Acup B$ where $D$ is the set of all divisors
    – Lozenges
    Aug 17 at 6:48
















  • Let $A$ be the set of all divisors $a$ of $20!$ such that $a<sqrt20!$. Define $B$ in a similar way. Now $D=Acup B$ where $D$ is the set of all divisors
    – Lozenges
    Aug 17 at 6:48















Let $A$ be the set of all divisors $a$ of $20!$ such that $a<sqrt20!$. Define $B$ in a similar way. Now $D=Acup B$ where $D$ is the set of all divisors
– Lozenges
Aug 17 at 6:48




Let $A$ be the set of all divisors $a$ of $20!$ such that $a<sqrt20!$. Define $B$ in a similar way. Now $D=Acup B$ where $D$ is the set of all divisors
– Lozenges
Aug 17 at 6:48










1 Answer
1






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up vote
4
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Hint: $a<b$ and they have no common prime factors. Now make a list of all the prime factors of $20!$....






share|cite|improve this answer




















  • got till that, but how do I proceed?
    – redx
    Aug 17 at 5:08










  • @redx try counting?
    – Stella Biderman
    Aug 17 at 5:08










  • do I have to use the highest powers of the prime factors?
    – redx
    Aug 17 at 5:09










  • @redx I don’t know what that question means.
    – Stella Biderman
    Aug 17 at 5:10






  • 1




    Choose a subset of the primes to be potential factors of $a$. How many ways to do that? The rest would be factors of $b$. Half the partitions will work because $a lt b$. We don't need to compute any of the $a$s or $b$s as we were just asked how many there are.
    – Ross Millikan
    Aug 17 at 14:26










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote













Hint: $a<b$ and they have no common prime factors. Now make a list of all the prime factors of $20!$....






share|cite|improve this answer




















  • got till that, but how do I proceed?
    – redx
    Aug 17 at 5:08










  • @redx try counting?
    – Stella Biderman
    Aug 17 at 5:08










  • do I have to use the highest powers of the prime factors?
    – redx
    Aug 17 at 5:09










  • @redx I don’t know what that question means.
    – Stella Biderman
    Aug 17 at 5:10






  • 1




    Choose a subset of the primes to be potential factors of $a$. How many ways to do that? The rest would be factors of $b$. Half the partitions will work because $a lt b$. We don't need to compute any of the $a$s or $b$s as we were just asked how many there are.
    – Ross Millikan
    Aug 17 at 14:26














up vote
4
down vote













Hint: $a<b$ and they have no common prime factors. Now make a list of all the prime factors of $20!$....






share|cite|improve this answer




















  • got till that, but how do I proceed?
    – redx
    Aug 17 at 5:08










  • @redx try counting?
    – Stella Biderman
    Aug 17 at 5:08










  • do I have to use the highest powers of the prime factors?
    – redx
    Aug 17 at 5:09










  • @redx I don’t know what that question means.
    – Stella Biderman
    Aug 17 at 5:10






  • 1




    Choose a subset of the primes to be potential factors of $a$. How many ways to do that? The rest would be factors of $b$. Half the partitions will work because $a lt b$. We don't need to compute any of the $a$s or $b$s as we were just asked how many there are.
    – Ross Millikan
    Aug 17 at 14:26












up vote
4
down vote










up vote
4
down vote









Hint: $a<b$ and they have no common prime factors. Now make a list of all the prime factors of $20!$....






share|cite|improve this answer












Hint: $a<b$ and they have no common prime factors. Now make a list of all the prime factors of $20!$....







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 17 at 5:02









Stella Biderman

26.1k63075




26.1k63075











  • got till that, but how do I proceed?
    – redx
    Aug 17 at 5:08










  • @redx try counting?
    – Stella Biderman
    Aug 17 at 5:08










  • do I have to use the highest powers of the prime factors?
    – redx
    Aug 17 at 5:09










  • @redx I don’t know what that question means.
    – Stella Biderman
    Aug 17 at 5:10






  • 1




    Choose a subset of the primes to be potential factors of $a$. How many ways to do that? The rest would be factors of $b$. Half the partitions will work because $a lt b$. We don't need to compute any of the $a$s or $b$s as we were just asked how many there are.
    – Ross Millikan
    Aug 17 at 14:26
















  • got till that, but how do I proceed?
    – redx
    Aug 17 at 5:08










  • @redx try counting?
    – Stella Biderman
    Aug 17 at 5:08










  • do I have to use the highest powers of the prime factors?
    – redx
    Aug 17 at 5:09










  • @redx I don’t know what that question means.
    – Stella Biderman
    Aug 17 at 5:10






  • 1




    Choose a subset of the primes to be potential factors of $a$. How many ways to do that? The rest would be factors of $b$. Half the partitions will work because $a lt b$. We don't need to compute any of the $a$s or $b$s as we were just asked how many there are.
    – Ross Millikan
    Aug 17 at 14:26















got till that, but how do I proceed?
– redx
Aug 17 at 5:08




got till that, but how do I proceed?
– redx
Aug 17 at 5:08












@redx try counting?
– Stella Biderman
Aug 17 at 5:08




@redx try counting?
– Stella Biderman
Aug 17 at 5:08












do I have to use the highest powers of the prime factors?
– redx
Aug 17 at 5:09




do I have to use the highest powers of the prime factors?
– redx
Aug 17 at 5:09












@redx I don’t know what that question means.
– Stella Biderman
Aug 17 at 5:10




@redx I don’t know what that question means.
– Stella Biderman
Aug 17 at 5:10




1




1




Choose a subset of the primes to be potential factors of $a$. How many ways to do that? The rest would be factors of $b$. Half the partitions will work because $a lt b$. We don't need to compute any of the $a$s or $b$s as we were just asked how many there are.
– Ross Millikan
Aug 17 at 14:26




Choose a subset of the primes to be potential factors of $a$. How many ways to do that? The rest would be factors of $b$. Half the partitions will work because $a lt b$. We don't need to compute any of the $a$s or $b$s as we were just asked how many there are.
– Ross Millikan
Aug 17 at 14:26












 

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