Given GCD of two numbers is 42 and their product is 15876. How many possible sets of members can be found?

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Given GCD of two numbers is 42 and their product is 15876. How many possible sets of numbers can be found?



I have no idea. I can only evaluate the lcm. Don't know how to get the answers.







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    up vote
    1
    down vote

    favorite












    Given GCD of two numbers is 42 and their product is 15876. How many possible sets of numbers can be found?



    I have no idea. I can only evaluate the lcm. Don't know how to get the answers.







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Given GCD of two numbers is 42 and their product is 15876. How many possible sets of numbers can be found?



      I have no idea. I can only evaluate the lcm. Don't know how to get the answers.







      share|cite|improve this question












      Given GCD of two numbers is 42 and their product is 15876. How many possible sets of numbers can be found?



      I have no idea. I can only evaluate the lcm. Don't know how to get the answers.









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 22 at 9:18









      user1942348

      1,3501625




      1,3501625




















          2 Answers
          2






          active

          oldest

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          up vote
          7
          down vote



          accepted










          You have two numbers that are divisible by $42$. We can write these as $42a$ and $42b$ for $a, b$ coprime (can you see why?). Their product is $15876$, so we have
          $$
          42^2ab = 15876 iff ab = 9.
          $$
          It it now enough to determine which $a, b$ satisfy $ab = 9$, where $a$ and $b$ are coprime.






          share|cite|improve this answer





























            up vote
            2
            down vote













            Use $gcd(x,y) times operatornamelcm(x,y) = xy$



            beginarrayrrr
            xy &= &15876 \
            operatornamelcm(x,y) &= &378 \
            hline
            gcd(x,y) &= &42
            endarray



            Assume $x < y$. Use $gcd(p^a, p^b) = p^min(a,b)$ and
            $operatornamelcm(p^a, p^b) = p^max(a,b)$ when $p$ is a prime number.



            beginarrayrcr
            gcd(x,y) &= &42 & 2^1 & 3^1 & 7^1 \
            operatornamelcm(x,y) &= &378 & 2^1 & 3^3 & 7^1 \
            xy &= &15876 & 2^2 & 3^4 & 7^2 \
            hline
            x &= &42 & 2^1 & 3^1 & 7^1 \
            y &= &378 & 2^1 & 3^3 & 7^1 \
            endarray






            share|cite|improve this answer






















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              7
              down vote



              accepted










              You have two numbers that are divisible by $42$. We can write these as $42a$ and $42b$ for $a, b$ coprime (can you see why?). Their product is $15876$, so we have
              $$
              42^2ab = 15876 iff ab = 9.
              $$
              It it now enough to determine which $a, b$ satisfy $ab = 9$, where $a$ and $b$ are coprime.






              share|cite|improve this answer


























                up vote
                7
                down vote



                accepted










                You have two numbers that are divisible by $42$. We can write these as $42a$ and $42b$ for $a, b$ coprime (can you see why?). Their product is $15876$, so we have
                $$
                42^2ab = 15876 iff ab = 9.
                $$
                It it now enough to determine which $a, b$ satisfy $ab = 9$, where $a$ and $b$ are coprime.






                share|cite|improve this answer
























                  up vote
                  7
                  down vote



                  accepted







                  up vote
                  7
                  down vote



                  accepted






                  You have two numbers that are divisible by $42$. We can write these as $42a$ and $42b$ for $a, b$ coprime (can you see why?). Their product is $15876$, so we have
                  $$
                  42^2ab = 15876 iff ab = 9.
                  $$
                  It it now enough to determine which $a, b$ satisfy $ab = 9$, where $a$ and $b$ are coprime.






                  share|cite|improve this answer














                  You have two numbers that are divisible by $42$. We can write these as $42a$ and $42b$ for $a, b$ coprime (can you see why?). Their product is $15876$, so we have
                  $$
                  42^2ab = 15876 iff ab = 9.
                  $$
                  It it now enough to determine which $a, b$ satisfy $ab = 9$, where $a$ and $b$ are coprime.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 22 at 9:27

























                  answered Aug 22 at 9:22









                  Bill Wallis

                  2,2361826




                  2,2361826




















                      up vote
                      2
                      down vote













                      Use $gcd(x,y) times operatornamelcm(x,y) = xy$



                      beginarrayrrr
                      xy &= &15876 \
                      operatornamelcm(x,y) &= &378 \
                      hline
                      gcd(x,y) &= &42
                      endarray



                      Assume $x < y$. Use $gcd(p^a, p^b) = p^min(a,b)$ and
                      $operatornamelcm(p^a, p^b) = p^max(a,b)$ when $p$ is a prime number.



                      beginarrayrcr
                      gcd(x,y) &= &42 & 2^1 & 3^1 & 7^1 \
                      operatornamelcm(x,y) &= &378 & 2^1 & 3^3 & 7^1 \
                      xy &= &15876 & 2^2 & 3^4 & 7^2 \
                      hline
                      x &= &42 & 2^1 & 3^1 & 7^1 \
                      y &= &378 & 2^1 & 3^3 & 7^1 \
                      endarray






                      share|cite|improve this answer


























                        up vote
                        2
                        down vote













                        Use $gcd(x,y) times operatornamelcm(x,y) = xy$



                        beginarrayrrr
                        xy &= &15876 \
                        operatornamelcm(x,y) &= &378 \
                        hline
                        gcd(x,y) &= &42
                        endarray



                        Assume $x < y$. Use $gcd(p^a, p^b) = p^min(a,b)$ and
                        $operatornamelcm(p^a, p^b) = p^max(a,b)$ when $p$ is a prime number.



                        beginarrayrcr
                        gcd(x,y) &= &42 & 2^1 & 3^1 & 7^1 \
                        operatornamelcm(x,y) &= &378 & 2^1 & 3^3 & 7^1 \
                        xy &= &15876 & 2^2 & 3^4 & 7^2 \
                        hline
                        x &= &42 & 2^1 & 3^1 & 7^1 \
                        y &= &378 & 2^1 & 3^3 & 7^1 \
                        endarray






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          Use $gcd(x,y) times operatornamelcm(x,y) = xy$



                          beginarrayrrr
                          xy &= &15876 \
                          operatornamelcm(x,y) &= &378 \
                          hline
                          gcd(x,y) &= &42
                          endarray



                          Assume $x < y$. Use $gcd(p^a, p^b) = p^min(a,b)$ and
                          $operatornamelcm(p^a, p^b) = p^max(a,b)$ when $p$ is a prime number.



                          beginarrayrcr
                          gcd(x,y) &= &42 & 2^1 & 3^1 & 7^1 \
                          operatornamelcm(x,y) &= &378 & 2^1 & 3^3 & 7^1 \
                          xy &= &15876 & 2^2 & 3^4 & 7^2 \
                          hline
                          x &= &42 & 2^1 & 3^1 & 7^1 \
                          y &= &378 & 2^1 & 3^3 & 7^1 \
                          endarray






                          share|cite|improve this answer














                          Use $gcd(x,y) times operatornamelcm(x,y) = xy$



                          beginarrayrrr
                          xy &= &15876 \
                          operatornamelcm(x,y) &= &378 \
                          hline
                          gcd(x,y) &= &42
                          endarray



                          Assume $x < y$. Use $gcd(p^a, p^b) = p^min(a,b)$ and
                          $operatornamelcm(p^a, p^b) = p^max(a,b)$ when $p$ is a prime number.



                          beginarrayrcr
                          gcd(x,y) &= &42 & 2^1 & 3^1 & 7^1 \
                          operatornamelcm(x,y) &= &378 & 2^1 & 3^3 & 7^1 \
                          xy &= &15876 & 2^2 & 3^4 & 7^2 \
                          hline
                          x &= &42 & 2^1 & 3^1 & 7^1 \
                          y &= &378 & 2^1 & 3^3 & 7^1 \
                          endarray







                          share|cite|improve this answer














                          share|cite|improve this answer



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                          edited Aug 22 at 14:48

























                          answered Aug 22 at 14:41









                          steven gregory

                          16.7k22155




                          16.7k22155






















                               

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