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The lcm of $p,q$ is $r^2t^4s^2$ where $p,q,r,s,t$ are natural numbers. Find the number of ordered pairs $(p,q)$.

-1

$begingroup$

The lcm of $p,q$ is $r^2t^4s^2$ where $p,q,r,s,t$ are natural numbers and $r,s,t$ are primes. Find the number of ordered pairs $(p,q)$.

I consider the set $r,r,t,t,t,t,s,s$ .I think the number of subsets of the set is the answer which is $2^8=256$.I am not sure about my approach.Can anyone solve this ?
Thank you in advance!

Edit: Suppose $r,s,t$ are fixed primes!

N.E.: It is a problem from the chapter 'elementary-set theory' .So you can this problem as a set-theoretic problem

combinatorics elementary-set-theory

share|cite|improve this question

edited Apr 2 at 3:18

Sufaid Saleel

asked Apr 2 at 2:25

Sufaid SaleelSufaid Saleel

1,766929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are $r, s$ and $t$ prime?
  $endgroup$
  – Mike
  Apr 2 at 2:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah they are prime
  $endgroup$
  – Sufaid Saleel
  Apr 2 at 2:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry about my answer; I wasn't quite right.
  $endgroup$
  – Mike
  Apr 2 at 2:56
  

  
  
  
  

add a comment | 

-1

$begingroup$

The lcm of $p,q$ is $r^2t^4s^2$ where $p,q,r,s,t$ are natural numbers and $r,s,t$ are primes. Find the number of ordered pairs $(p,q)$.

I consider the set $r,r,t,t,t,t,s,s$ .I think the number of subsets of the set is the answer which is $2^8=256$.I am not sure about my approach.Can anyone solve this ?
Thank you in advance!

Edit: Suppose $r,s,t$ are fixed primes!

N.E.: It is a problem from the chapter 'elementary-set theory' .So you can this problem as a set-theoretic problem

combinatorics elementary-set-theory

share|cite|improve this question

edited Apr 2 at 3:18

Sufaid Saleel

asked Apr 2 at 2:25

Sufaid SaleelSufaid Saleel

1,766929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are $r, s$ and $t$ prime?
  $endgroup$
  – Mike
  Apr 2 at 2:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah they are prime
  $endgroup$
  – Sufaid Saleel
  Apr 2 at 2:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry about my answer; I wasn't quite right.
  $endgroup$
  – Mike
  Apr 2 at 2:56
  

  
  
  
  

add a comment | 

$begingroup$

The lcm of $p,q$ is $r^2t^4s^2$ where $p,q,r,s,t$ are natural numbers and $r,s,t$ are primes. Find the number of ordered pairs $(p,q)$.

I consider the set $r,r,t,t,t,t,s,s$ .I think the number of subsets of the set is the answer which is $2^8=256$.I am not sure about my approach.Can anyone solve this ?
Thank you in advance!

Edit: Suppose $r,s,t$ are fixed primes!

N.E.: It is a problem from the chapter 'elementary-set theory' .So you can this problem as a set-theoretic problem

combinatorics elementary-set-theory

share|cite|improve this question

edited Apr 2 at 3:18

Sufaid Saleel

asked Apr 2 at 2:25

Sufaid SaleelSufaid Saleel

1,766929

$endgroup$

share|cite|improve this question

edited Apr 2 at 3:18

Sufaid Saleel

asked Apr 2 at 2:25

Sufaid SaleelSufaid Saleel

1,766929

share|cite|improve this question

edited Apr 2 at 3:18

Sufaid Saleel

asked Apr 2 at 2:25

Sufaid SaleelSufaid Saleel

1,766929

asked Apr 2 at 2:25

Sufaid SaleelSufaid Saleel

1,766929

asked Apr 2 at 2:25

Sufaid SaleelSufaid Saleel

1,766929

1 Answer
1

active

oldest

votes

1

$begingroup$

There is a countable infinity. Find any single pair $(p,q)$ that satisfies the requirement. I would just choose $p=1, q=r^2s^4t^2$ because it is easy to find. You can choose $r,s,t$ to be any primes you want.

For fixed $r,s,t$ you need each of $p=r^as^bt^c,q=r^ds^et^f$ with $max(a,d)=2$ and so on. How many choices for $a,d$? Count them and multiply.

share|cite|improve this answer

edited Apr 2 at 2:43

answered Apr 2 at 2:36

Ross MillikanRoss Millikan

302k24201375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I asked for the number of such ordered pairs!
  $endgroup$
  – Sufaid Saleel
  Apr 2 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is in the first sentence. There are as many as the naturals. The set of primes is infinite.
  $endgroup$
  – Ross Millikan
  Apr 2 at 2:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Suppose $r,s,t$ are fixed primes!
  $endgroup$
  – Sufaid Saleel
  Apr 2 at 2:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have made two major changes to the question. Please think about the question and make sure you have it right before posting.
  $endgroup$
  – Ross Millikan
  Apr 2 at 2:40
  

  
  
  
  

add a comment | 

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1

$begingroup$

There is a countable infinity. Find any single pair $(p,q)$ that satisfies the requirement. I would just choose $p=1, q=r^2s^4t^2$ because it is easy to find. You can choose $r,s,t$ to be any primes you want.

For fixed $r,s,t$ you need each of $p=r^as^bt^c,q=r^ds^et^f$ with $max(a,d)=2$ and so on. How many choices for $a,d$? Count them and multiply.

share|cite|improve this answer

edited Apr 2 at 2:43

answered Apr 2 at 2:36

Ross MillikanRoss Millikan

302k24201375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I asked for the number of such ordered pairs!
  $endgroup$
  – Sufaid Saleel
  Apr 2 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is in the first sentence. There are as many as the naturals. The set of primes is infinite.
  $endgroup$
  – Ross Millikan
  Apr 2 at 2:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Suppose $r,s,t$ are fixed primes!
  $endgroup$
  – Sufaid Saleel
  Apr 2 at 2:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have made two major changes to the question. Please think about the question and make sure you have it right before posting.
  $endgroup$
  – Ross Millikan
  Apr 2 at 2:40
  

  
  
  
  

add a comment | 

1

$begingroup$

There is a countable infinity. Find any single pair $(p,q)$ that satisfies the requirement. I would just choose $p=1, q=r^2s^4t^2$ because it is easy to find. You can choose $r,s,t$ to be any primes you want.

For fixed $r,s,t$ you need each of $p=r^as^bt^c,q=r^ds^et^f$ with $max(a,d)=2$ and so on. How many choices for $a,d$? Count them and multiply.

share|cite|improve this answer

edited Apr 2 at 2:43

answered Apr 2 at 2:36

Ross MillikanRoss Millikan

302k24201375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I asked for the number of such ordered pairs!
  $endgroup$
  – Sufaid Saleel
  Apr 2 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is in the first sentence. There are as many as the naturals. The set of primes is infinite.
  $endgroup$
  – Ross Millikan
  Apr 2 at 2:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Suppose $r,s,t$ are fixed primes!
  $endgroup$
  – Sufaid Saleel
  Apr 2 at 2:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have made two major changes to the question. Please think about the question and make sure you have it right before posting.
  $endgroup$
  – Ross Millikan
  Apr 2 at 2:40
  

  
  
  
  

add a comment | 

$begingroup$

There is a countable infinity. Find any single pair $(p,q)$ that satisfies the requirement. I would just choose $p=1, q=r^2s^4t^2$ because it is easy to find. You can choose $r,s,t$ to be any primes you want.

For fixed $r,s,t$ you need each of $p=r^as^bt^c,q=r^ds^et^f$ with $max(a,d)=2$ and so on. How many choices for $a,d$? Count them and multiply.

share|cite|improve this answer

edited Apr 2 at 2:43

answered Apr 2 at 2:36

Ross MillikanRoss Millikan

302k24201375

$endgroup$

share|cite|improve this answer

edited Apr 2 at 2:43

answered Apr 2 at 2:36

Ross MillikanRoss Millikan

302k24201375

answered Apr 2 at 2:36

Ross MillikanRoss Millikan

302k24201375

answered Apr 2 at 2:36

Ross MillikanRoss Millikan

302k24201375