Find the number of ordered triplets












0












$begingroup$


It is an olympiad problem.
Find all ordered triples of positive integers(a,b,c),
Such that 1/a+1/b+1/c=3/4



Till now i got only 1 solutions, but i expect there are more than that.



I brought 4 to the LHS and got 4/a+4/b+4/c=3, it is trivial though, and hence a=b=c=4.
Hence from the above one, i got 1 solution.
I expect there are many more but could not find it, ido not know what to do next. Please help!



Edit: i found there must be 25 solutions in all






number-theory contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    I assume these must be integers?
    $endgroup$
    – Matt Samuel
    Jan 19 at 13:52










  • $begingroup$
    Don't forget solutions with negative terms, like $(1,-2,4)$
    $endgroup$
    – lulu
    Jan 19 at 13:53










  • $begingroup$
    These are positive integers
    $endgroup$
    – user636268
    Jan 19 at 13:54






  • 1




    $begingroup$
    Please edit the question to include all the requirements you have in mind.
    $endgroup$
    – lulu
    Jan 19 at 13:55










  • $begingroup$
    Yeah you are right
    $endgroup$
    – user636268
    Jan 19 at 13:55
















0












$begingroup$


It is an olympiad problem.
Find all ordered triples of positive integers(a,b,c),
Such that 1/a+1/b+1/c=3/4



Till now i got only 1 solutions, but i expect there are more than that.



I brought 4 to the LHS and got 4/a+4/b+4/c=3, it is trivial though, and hence a=b=c=4.
Hence from the above one, i got 1 solution.
I expect there are many more but could not find it, ido not know what to do next. Please help!



Edit: i found there must be 25 solutions in all






number-theory contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    I assume these must be integers?
    $endgroup$
    – Matt Samuel
    Jan 19 at 13:52










  • $begingroup$
    Don't forget solutions with negative terms, like $(1,-2,4)$
    $endgroup$
    – lulu
    Jan 19 at 13:53










  • $begingroup$
    These are positive integers
    $endgroup$
    – user636268
    Jan 19 at 13:54






  • 1




    $begingroup$
    Please edit the question to include all the requirements you have in mind.
    $endgroup$
    – lulu
    Jan 19 at 13:55










  • $begingroup$
    Yeah you are right
    $endgroup$
    – user636268
    Jan 19 at 13:55














0












0








0





$begingroup$


It is an olympiad problem.
Find all ordered triples of positive integers(a,b,c),
Such that 1/a+1/b+1/c=3/4



Till now i got only 1 solutions, but i expect there are more than that.



I brought 4 to the LHS and got 4/a+4/b+4/c=3, it is trivial though, and hence a=b=c=4.
Hence from the above one, i got 1 solution.
I expect there are many more but could not find it, ido not know what to do next. Please help!



Edit: i found there must be 25 solutions in all






number-theory contest-math







share|cite|improve this question











$endgroup$




It is an olympiad problem.
Find all ordered triples of positive integers(a,b,c),
Such that 1/a+1/b+1/c=3/4



Till now i got only 1 solutions, but i expect there are more than that.



I brought 4 to the LHS and got 4/a+4/b+4/c=3, it is trivial though, and hence a=b=c=4.
Hence from the above one, i got 1 solution.
I expect there are many more but could not find it, ido not know what to do next. Please help!



Edit: i found there must be 25 solutions in all






number-theory contest-math




number-theory contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 13:56

























asked Jan 19 at 13:42







user636268



















  • $begingroup$
    I assume these must be integers?
    $endgroup$
    – Matt Samuel
    Jan 19 at 13:52










  • $begingroup$
    Don't forget solutions with negative terms, like $(1,-2,4)$
    $endgroup$
    – lulu
    Jan 19 at 13:53










  • $begingroup$
    These are positive integers
    $endgroup$
    – user636268
    Jan 19 at 13:54






  • 1




    $begingroup$
    Please edit the question to include all the requirements you have in mind.
    $endgroup$
    – lulu
    Jan 19 at 13:55










  • $begingroup$
    Yeah you are right
    $endgroup$
    – user636268
    Jan 19 at 13:55


















  • $begingroup$
    I assume these must be integers?
    $endgroup$
    – Matt Samuel
    Jan 19 at 13:52










  • $begingroup$
    Don't forget solutions with negative terms, like $(1,-2,4)$
    $endgroup$
    – lulu
    Jan 19 at 13:53










  • $begingroup$
    These are positive integers
    $endgroup$
    – user636268
    Jan 19 at 13:54






  • 1




    $begingroup$
    Please edit the question to include all the requirements you have in mind.
    $endgroup$
    – lulu
    Jan 19 at 13:55










  • $begingroup$
    Yeah you are right
    $endgroup$
    – user636268
    Jan 19 at 13:55
















$begingroup$
I assume these must be integers?
$endgroup$
– Matt Samuel
Jan 19 at 13:52




$begingroup$
I assume these must be integers?
$endgroup$
– Matt Samuel
Jan 19 at 13:52












$begingroup$
Don't forget solutions with negative terms, like $(1,-2,4)$
$endgroup$
– lulu
Jan 19 at 13:53




$begingroup$
Don't forget solutions with negative terms, like $(1,-2,4)$
$endgroup$
– lulu
Jan 19 at 13:53












$begingroup$
These are positive integers
$endgroup$
– user636268
Jan 19 at 13:54




$begingroup$
These are positive integers
$endgroup$
– user636268
Jan 19 at 13:54




1




1




$begingroup$
Please edit the question to include all the requirements you have in mind.
$endgroup$
– lulu
Jan 19 at 13:55




$begingroup$
Please edit the question to include all the requirements you have in mind.
$endgroup$
– lulu
Jan 19 at 13:55












$begingroup$
Yeah you are right
$endgroup$
– user636268
Jan 19 at 13:55




$begingroup$
Yeah you are right
$endgroup$
– user636268
Jan 19 at 13:55










2 Answers
2






active

oldest

votes


















0












$begingroup$

If the smallest denominator is $4$, then they are all $4$, giving $frac{1}{4} + frac{1}{4} + frac{1}{4}$.



If the smallest denominator is $3$, then $frac{1}{b} + frac{1}{c} = frac{5}{12}$. Either $b$ or $c$ must be less than or equal to $4$. If $b=4$, $c=6$, giving $frac{1}{3}+ frac{1}{4} + frac{1}{6}$. If $b=3$, $c=12$, giving $frac{1}{3} + frac{1}{3} + frac{1}{12}$.



If the smallest denominator is $2$, then $frac{1}{b} + frac{1}{c} = frac{1}{4}$. Either $b$ or $c$ must be less than or equal to $8$. This gives several more solutions:
$${a,b,c} = {2,5,20}, {2,6,12}, {2,8,8}.$$



From there, you just need to figure out how many orderings of each triplet there are.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your solution is correct!
    $endgroup$
    – user636268
    Jan 19 at 14:05



















0












$begingroup$

Assume without loss of generality that $2le ale ble c$. Then
$$
frac{3}{a}gefrac{1}{a}+frac{1}{b}+frac{1}{c}=frac{3}{4}.
$$ Hence $2le ale 4$. If $a=2$, then $frac{ 1}{b}+frac{1}{c}=frac{1}{4}$. Since $frac{2}{b}ge frac{1}{b}+frac{1}{c}$, we get $2le ble 8$. By Exhaustive search, we get $(b,c)=(5,20),(6,12),(8,8)$.



If $a=3$, then $frac{1}{b}+frac{1}{c}=frac{5}{12}$ and $3le ble 24/5$. In the same way, this gives $(b,c)=(3,12),(4,6)$.



If $a=4$, then $frac{1}{b}+frac{1}{c}=frac{1}{2}$ and $4le ble 4$. It is only possible for $(b,c)=(4,4)$.



These are all solutions of the equation with $ale ble c$. And any solution $(a,b,c)$ is a permutation of one of these.






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









    0












    $begingroup$

    If the smallest denominator is $4$, then they are all $4$, giving $frac{1}{4} + frac{1}{4} + frac{1}{4}$.



    If the smallest denominator is $3$, then $frac{1}{b} + frac{1}{c} = frac{5}{12}$. Either $b$ or $c$ must be less than or equal to $4$. If $b=4$, $c=6$, giving $frac{1}{3}+ frac{1}{4} + frac{1}{6}$. If $b=3$, $c=12$, giving $frac{1}{3} + frac{1}{3} + frac{1}{12}$.



    If the smallest denominator is $2$, then $frac{1}{b} + frac{1}{c} = frac{1}{4}$. Either $b$ or $c$ must be less than or equal to $8$. This gives several more solutions:
    $${a,b,c} = {2,5,20}, {2,6,12}, {2,8,8}.$$



    From there, you just need to figure out how many orderings of each triplet there are.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Your solution is correct!
      $endgroup$
      – user636268
      Jan 19 at 14:05
















    0












    $begingroup$

    If the smallest denominator is $4$, then they are all $4$, giving $frac{1}{4} + frac{1}{4} + frac{1}{4}$.



    If the smallest denominator is $3$, then $frac{1}{b} + frac{1}{c} = frac{5}{12}$. Either $b$ or $c$ must be less than or equal to $4$. If $b=4$, $c=6$, giving $frac{1}{3}+ frac{1}{4} + frac{1}{6}$. If $b=3$, $c=12$, giving $frac{1}{3} + frac{1}{3} + frac{1}{12}$.



    If the smallest denominator is $2$, then $frac{1}{b} + frac{1}{c} = frac{1}{4}$. Either $b$ or $c$ must be less than or equal to $8$. This gives several more solutions:
    $${a,b,c} = {2,5,20}, {2,6,12}, {2,8,8}.$$



    From there, you just need to figure out how many orderings of each triplet there are.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Your solution is correct!
      $endgroup$
      – user636268
      Jan 19 at 14:05














    0












    0








    0





    $begingroup$

    If the smallest denominator is $4$, then they are all $4$, giving $frac{1}{4} + frac{1}{4} + frac{1}{4}$.



    If the smallest denominator is $3$, then $frac{1}{b} + frac{1}{c} = frac{5}{12}$. Either $b$ or $c$ must be less than or equal to $4$. If $b=4$, $c=6$, giving $frac{1}{3}+ frac{1}{4} + frac{1}{6}$. If $b=3$, $c=12$, giving $frac{1}{3} + frac{1}{3} + frac{1}{12}$.



    If the smallest denominator is $2$, then $frac{1}{b} + frac{1}{c} = frac{1}{4}$. Either $b$ or $c$ must be less than or equal to $8$. This gives several more solutions:
    $${a,b,c} = {2,5,20}, {2,6,12}, {2,8,8}.$$



    From there, you just need to figure out how many orderings of each triplet there are.






    share|cite|improve this answer









    $endgroup$



    If the smallest denominator is $4$, then they are all $4$, giving $frac{1}{4} + frac{1}{4} + frac{1}{4}$.



    If the smallest denominator is $3$, then $frac{1}{b} + frac{1}{c} = frac{5}{12}$. Either $b$ or $c$ must be less than or equal to $4$. If $b=4$, $c=6$, giving $frac{1}{3}+ frac{1}{4} + frac{1}{6}$. If $b=3$, $c=12$, giving $frac{1}{3} + frac{1}{3} + frac{1}{12}$.



    If the smallest denominator is $2$, then $frac{1}{b} + frac{1}{c} = frac{1}{4}$. Either $b$ or $c$ must be less than or equal to $8$. This gives several more solutions:
    $${a,b,c} = {2,5,20}, {2,6,12}, {2,8,8}.$$



    From there, you just need to figure out how many orderings of each triplet there are.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 19 at 13:59









    rogerlrogerl

    17.8k22747




    17.8k22747












    • $begingroup$
      Your solution is correct!
      $endgroup$
      – user636268
      Jan 19 at 14:05


















    • $begingroup$
      Your solution is correct!
      $endgroup$
      – user636268
      Jan 19 at 14:05
















    $begingroup$
    Your solution is correct!
    $endgroup$
    – user636268
    Jan 19 at 14:05




    $begingroup$
    Your solution is correct!
    $endgroup$
    – user636268
    Jan 19 at 14:05











    0












    $begingroup$

    Assume without loss of generality that $2le ale ble c$. Then
    $$
    frac{3}{a}gefrac{1}{a}+frac{1}{b}+frac{1}{c}=frac{3}{4}.
    $$     Hence $2le ale 4$. If $a=2$, then $frac{ 1}{b}+frac{1}{c}=frac{1}{4}$. Since $frac{2}{b}ge frac{1}{b}+frac{1}{c}$, we get $2le ble 8$. By Exhaustive search, we get $(b,c)=(5,20),(6,12),(8,8)$.



    If $a=3$, then $frac{1}{b}+frac{1}{c}=frac{5}{12}$ and $3le ble 24/5$. In the same way, this gives $(b,c)=(3,12),(4,6)$.



    If $a=4$, then $frac{1}{b}+frac{1}{c}=frac{1}{2}$ and $4le ble 4$. It is only possible for $(b,c)=(4,4)$.



    These are all solutions of the equation with $ale ble c$. And any solution $(a,b,c)$ is a permutation of one of these.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Assume without loss of generality that $2le ale ble c$. Then
      $$
      frac{3}{a}gefrac{1}{a}+frac{1}{b}+frac{1}{c}=frac{3}{4}.
      $$       Hence $2le ale 4$. If $a=2$, then $frac{ 1}{b}+frac{1}{c}=frac{1}{4}$. Since $frac{2}{b}ge frac{1}{b}+frac{1}{c}$, we get $2le ble 8$. By Exhaustive search, we get $(b,c)=(5,20),(6,12),(8,8)$.



      If $a=3$, then $frac{1}{b}+frac{1}{c}=frac{5}{12}$ and $3le ble 24/5$. In the same way, this gives $(b,c)=(3,12),(4,6)$.



      If $a=4$, then $frac{1}{b}+frac{1}{c}=frac{1}{2}$ and $4le ble 4$. It is only possible for $(b,c)=(4,4)$.



      These are all solutions of the equation with $ale ble c$. And any solution $(a,b,c)$ is a permutation of one of these.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Assume without loss of generality that $2le ale ble c$. Then
        $$
        frac{3}{a}gefrac{1}{a}+frac{1}{b}+frac{1}{c}=frac{3}{4}.
        $$         Hence $2le ale 4$. If $a=2$, then $frac{ 1}{b}+frac{1}{c}=frac{1}{4}$. Since $frac{2}{b}ge frac{1}{b}+frac{1}{c}$, we get $2le ble 8$. By Exhaustive search, we get $(b,c)=(5,20),(6,12),(8,8)$.



        If $a=3$, then $frac{1}{b}+frac{1}{c}=frac{5}{12}$ and $3le ble 24/5$. In the same way, this gives $(b,c)=(3,12),(4,6)$.



        If $a=4$, then $frac{1}{b}+frac{1}{c}=frac{1}{2}$ and $4le ble 4$. It is only possible for $(b,c)=(4,4)$.



        These are all solutions of the equation with $ale ble c$. And any solution $(a,b,c)$ is a permutation of one of these.






        share|cite|improve this answer









        $endgroup$



        Assume without loss of generality that $2le ale ble c$. Then
        $$
        frac{3}{a}gefrac{1}{a}+frac{1}{b}+frac{1}{c}=frac{3}{4}.
        $$         Hence $2le ale 4$. If $a=2$, then $frac{ 1}{b}+frac{1}{c}=frac{1}{4}$. Since $frac{2}{b}ge frac{1}{b}+frac{1}{c}$, we get $2le ble 8$. By Exhaustive search, we get $(b,c)=(5,20),(6,12),(8,8)$.



        If $a=3$, then $frac{1}{b}+frac{1}{c}=frac{5}{12}$ and $3le ble 24/5$. In the same way, this gives $(b,c)=(3,12),(4,6)$.



        If $a=4$, then $frac{1}{b}+frac{1}{c}=frac{1}{2}$ and $4le ble 4$. It is only possible for $(b,c)=(4,4)$.



        These are all solutions of the equation with $ale ble c$. And any solution $(a,b,c)$ is a permutation of one of these.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 19 at 14:00









        SongSong

        13.6k633




        13.6k633






























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