How many reduced fractions a/b such that ab=20! and 0 < a/b < 1?
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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
add a comment |Â
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
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
â Lozenges
Aug 17 at 6:48
add a comment |Â
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
(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
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
â Lozenges
Aug 17 at 6:48
add a comment |Â
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
â 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
â 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
â Lozenges
Aug 17 at 6:48
add a comment |Â
1 Answer
1
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 $20!$....
got till that, but how do I proceed?
â redx
Aug 17 at 5:08
@redx try counting?
â Stella Biderman
Aug 17 at 5:08
do I have to use the highest powers of the prime factors?
â redx
Aug 17 at 5:09
@redx I donâÂÂt know what that question means.
â 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.
â Ross Millikan
Aug 17 at 14:26
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
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 $20!$....
got till that, but how do I proceed?
â redx
Aug 17 at 5:08
@redx try counting?
â Stella Biderman
Aug 17 at 5:08
do I have to use the highest powers of the prime factors?
â redx
Aug 17 at 5:09
@redx I donâÂÂt know what that question means.
â 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.
â Ross Millikan
Aug 17 at 14:26
 |Â
show 2 more comments
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 $20!$....
got till that, but how do I proceed?
â redx
Aug 17 at 5:08
@redx try counting?
â Stella Biderman
Aug 17 at 5:08
do I have to use the highest powers of the prime factors?
â redx
Aug 17 at 5:09
@redx I donâÂÂt know what that question means.
â 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.
â Ross Millikan
Aug 17 at 14:26
 |Â
show 2 more comments
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 $20!$....
Hint: $a<b$ and they have no common prime factors. Now make a list of all the prime factors of $20!$....
answered Aug 17 at 5:02
Stella Biderman
26.1k63075
26.1k63075
got till that, but how do I proceed?
â redx
Aug 17 at 5:08
@redx try counting?
â Stella Biderman
Aug 17 at 5:08
do I have to use the highest powers of the prime factors?
â redx
Aug 17 at 5:09
@redx I donâÂÂt know what that question means.
â 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.
â Ross Millikan
Aug 17 at 14:26
 |Â
show 2 more comments
got till that, but how do I proceed?
â redx
Aug 17 at 5:08
@redx try counting?
â Stella Biderman
Aug 17 at 5:08
do I have to use the highest powers of the prime factors?
â redx
Aug 17 at 5:09
@redx I donâÂÂt know what that question means.
â 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.
â Ross Millikan
Aug 17 at 14:26
got till that, but how do I proceed?
â redx
Aug 17 at 5:08
got till that, but how do I proceed?
â redx
Aug 17 at 5:08
@redx try counting?
â Stella Biderman
Aug 17 at 5:08
@redx try counting?
â Stella Biderman
Aug 17 at 5:08
do I have to use the highest powers of the prime factors?
â redx
Aug 17 at 5:09
do I have to use the highest powers of the prime factors?
â redx
Aug 17 at 5:09
@redx I donâÂÂt know what that question means.
â Stella Biderman
Aug 17 at 5:10
@redx I donâÂÂt know what that question means.
â 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.
â 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.
â Ross Millikan
Aug 17 at 14:26
 |Â
show 2 more comments
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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
â Lozenges
Aug 17 at 6:48