How to calculate the sum of the multiples of two fractions in order?












0












$begingroup$


$frac{a}{b_1}, frac{a}{b_2} = frac{1}{3}, frac{1}{5}$



$x_1 in {0, ..., b_1-1} = {0, 1, 2}$



$x_2 in {0, ..., b_2-1} = {0, 1, 2, 3, 4}$



$$frac{a_1}{b}, frac{a_2}{b} = frac{a_1b_2}{b_1b_2}, frac{a_2b_1}{b_1b_2} = frac{5}{15}, frac{3}{15}$$



$$frac{p}{q} = frac{a_1x_1}{b} + frac{a_2x_2}{b}$$



I've been struggling for a couple of days to find an equation that can calculate $frac{p}{q}$ in order from smallest to largest. Any help will be appreciated thank you.



The sums in order:



$$frac{5cdot0 + 3cdot0}{15} = frac{0}{15}, frac{5cdot0 + 3cdot1}{15} = frac{3}{15}, frac{5cdot1 + 3cdot0}{15} = frac{5}{15}, frac{5cdot0 + 3cdot2}{15} = frac{6}{15}$$



$$frac{5cdot1 + 3cdot1}{15} = frac{8}{15}, frac{5cdot0 + 3cdot3}{15} = frac{9}{15}, frac{5cdot2 + 3cdot0}{15} = frac{10}{15}, frac{5cdot1 + 3cdot2}{15} = frac{11}{15}$$



$$frac{5cdot0 + 3cdot4}{15} = frac{12}{15}, frac{5cdot2 + 3cdot1}{15} = frac{13}{15}, frac{5cdot1 + 3cdot3}{15} = frac{14}{15}, frac{5cdot2 + 3cdot2}{15} = frac{16}{15}$$



$$frac{5cdot1 + 3cdot4}{15} = frac{17}{15}, frac{5cdot2 + 3cdot3}{15} = frac{19}{15}, frac{5cdot2 + 3cdot4}{15} = frac{22}{15}$$






combinatorics fractions







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$endgroup$












  • $begingroup$
    Why don't you want to compute them all and then sort them? You can organize the computation us use the fact that subgroups are presorted.
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 15:08










  • $begingroup$
    @RossMillikan Because I'm not writing s computer program
    $endgroup$
    – user553185
    Dec 1 '18 at 15:09










  • $begingroup$
    Then you can just make a list for each $x_1$ going over the values of $x_2$, so you get $(0,3,6,9,12), (5,8,11,14,17), (10,13,16,19,22)$ and pick the smallest value off the front of one, getting $(0,3,5,6,8,9,10,11,12,13,14,16,17,19,22)$ Your denominators are constant, so you just need to compute the numerators.
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 15:38












  • $begingroup$
    @RossMillikan The goal is primarily to find an equation that can do it.
    $endgroup$
    – user553185
    Dec 1 '18 at 15:40






  • 1




    $begingroup$
    Do you want the denominator to be the least common multiple of 3 and 5? Or is it just there product?
    $endgroup$
    – Fareed AF
    Dec 3 '18 at 11:23


















0












$begingroup$


$frac{a}{b_1}, frac{a}{b_2} = frac{1}{3}, frac{1}{5}$



$x_1 in {0, ..., b_1-1} = {0, 1, 2}$



$x_2 in {0, ..., b_2-1} = {0, 1, 2, 3, 4}$



$$frac{a_1}{b}, frac{a_2}{b} = frac{a_1b_2}{b_1b_2}, frac{a_2b_1}{b_1b_2} = frac{5}{15}, frac{3}{15}$$



$$frac{p}{q} = frac{a_1x_1}{b} + frac{a_2x_2}{b}$$



I've been struggling for a couple of days to find an equation that can calculate $frac{p}{q}$ in order from smallest to largest. Any help will be appreciated thank you.



The sums in order:



$$frac{5cdot0 + 3cdot0}{15} = frac{0}{15}, frac{5cdot0 + 3cdot1}{15} = frac{3}{15}, frac{5cdot1 + 3cdot0}{15} = frac{5}{15}, frac{5cdot0 + 3cdot2}{15} = frac{6}{15}$$



$$frac{5cdot1 + 3cdot1}{15} = frac{8}{15}, frac{5cdot0 + 3cdot3}{15} = frac{9}{15}, frac{5cdot2 + 3cdot0}{15} = frac{10}{15}, frac{5cdot1 + 3cdot2}{15} = frac{11}{15}$$



$$frac{5cdot0 + 3cdot4}{15} = frac{12}{15}, frac{5cdot2 + 3cdot1}{15} = frac{13}{15}, frac{5cdot1 + 3cdot3}{15} = frac{14}{15}, frac{5cdot2 + 3cdot2}{15} = frac{16}{15}$$



$$frac{5cdot1 + 3cdot4}{15} = frac{17}{15}, frac{5cdot2 + 3cdot3}{15} = frac{19}{15}, frac{5cdot2 + 3cdot4}{15} = frac{22}{15}$$






combinatorics fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why don't you want to compute them all and then sort them? You can organize the computation us use the fact that subgroups are presorted.
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 15:08










  • $begingroup$
    @RossMillikan Because I'm not writing s computer program
    $endgroup$
    – user553185
    Dec 1 '18 at 15:09










  • $begingroup$
    Then you can just make a list for each $x_1$ going over the values of $x_2$, so you get $(0,3,6,9,12), (5,8,11,14,17), (10,13,16,19,22)$ and pick the smallest value off the front of one, getting $(0,3,5,6,8,9,10,11,12,13,14,16,17,19,22)$ Your denominators are constant, so you just need to compute the numerators.
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 15:38












  • $begingroup$
    @RossMillikan The goal is primarily to find an equation that can do it.
    $endgroup$
    – user553185
    Dec 1 '18 at 15:40






  • 1




    $begingroup$
    Do you want the denominator to be the least common multiple of 3 and 5? Or is it just there product?
    $endgroup$
    – Fareed AF
    Dec 3 '18 at 11:23
















0












0








0


1



$begingroup$


$frac{a}{b_1}, frac{a}{b_2} = frac{1}{3}, frac{1}{5}$



$x_1 in {0, ..., b_1-1} = {0, 1, 2}$



$x_2 in {0, ..., b_2-1} = {0, 1, 2, 3, 4}$



$$frac{a_1}{b}, frac{a_2}{b} = frac{a_1b_2}{b_1b_2}, frac{a_2b_1}{b_1b_2} = frac{5}{15}, frac{3}{15}$$



$$frac{p}{q} = frac{a_1x_1}{b} + frac{a_2x_2}{b}$$



I've been struggling for a couple of days to find an equation that can calculate $frac{p}{q}$ in order from smallest to largest. Any help will be appreciated thank you.



The sums in order:



$$frac{5cdot0 + 3cdot0}{15} = frac{0}{15}, frac{5cdot0 + 3cdot1}{15} = frac{3}{15}, frac{5cdot1 + 3cdot0}{15} = frac{5}{15}, frac{5cdot0 + 3cdot2}{15} = frac{6}{15}$$



$$frac{5cdot1 + 3cdot1}{15} = frac{8}{15}, frac{5cdot0 + 3cdot3}{15} = frac{9}{15}, frac{5cdot2 + 3cdot0}{15} = frac{10}{15}, frac{5cdot1 + 3cdot2}{15} = frac{11}{15}$$



$$frac{5cdot0 + 3cdot4}{15} = frac{12}{15}, frac{5cdot2 + 3cdot1}{15} = frac{13}{15}, frac{5cdot1 + 3cdot3}{15} = frac{14}{15}, frac{5cdot2 + 3cdot2}{15} = frac{16}{15}$$



$$frac{5cdot1 + 3cdot4}{15} = frac{17}{15}, frac{5cdot2 + 3cdot3}{15} = frac{19}{15}, frac{5cdot2 + 3cdot4}{15} = frac{22}{15}$$






combinatorics fractions







share|cite|improve this question











$endgroup$




$frac{a}{b_1}, frac{a}{b_2} = frac{1}{3}, frac{1}{5}$



$x_1 in {0, ..., b_1-1} = {0, 1, 2}$



$x_2 in {0, ..., b_2-1} = {0, 1, 2, 3, 4}$



$$frac{a_1}{b}, frac{a_2}{b} = frac{a_1b_2}{b_1b_2}, frac{a_2b_1}{b_1b_2} = frac{5}{15}, frac{3}{15}$$



$$frac{p}{q} = frac{a_1x_1}{b} + frac{a_2x_2}{b}$$



I've been struggling for a couple of days to find an equation that can calculate $frac{p}{q}$ in order from smallest to largest. Any help will be appreciated thank you.



The sums in order:



$$frac{5cdot0 + 3cdot0}{15} = frac{0}{15}, frac{5cdot0 + 3cdot1}{15} = frac{3}{15}, frac{5cdot1 + 3cdot0}{15} = frac{5}{15}, frac{5cdot0 + 3cdot2}{15} = frac{6}{15}$$



$$frac{5cdot1 + 3cdot1}{15} = frac{8}{15}, frac{5cdot0 + 3cdot3}{15} = frac{9}{15}, frac{5cdot2 + 3cdot0}{15} = frac{10}{15}, frac{5cdot1 + 3cdot2}{15} = frac{11}{15}$$



$$frac{5cdot0 + 3cdot4}{15} = frac{12}{15}, frac{5cdot2 + 3cdot1}{15} = frac{13}{15}, frac{5cdot1 + 3cdot3}{15} = frac{14}{15}, frac{5cdot2 + 3cdot2}{15} = frac{16}{15}$$



$$frac{5cdot1 + 3cdot4}{15} = frac{17}{15}, frac{5cdot2 + 3cdot3}{15} = frac{19}{15}, frac{5cdot2 + 3cdot4}{15} = frac{22}{15}$$






combinatorics fractions




combinatorics fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 22:57









Byte Commander

196211




196211










asked Dec 1 '18 at 14:34









user553185user553185

155




155












  • $begingroup$
    Why don't you want to compute them all and then sort them? You can organize the computation us use the fact that subgroups are presorted.
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 15:08










  • $begingroup$
    @RossMillikan Because I'm not writing s computer program
    $endgroup$
    – user553185
    Dec 1 '18 at 15:09










  • $begingroup$
    Then you can just make a list for each $x_1$ going over the values of $x_2$, so you get $(0,3,6,9,12), (5,8,11,14,17), (10,13,16,19,22)$ and pick the smallest value off the front of one, getting $(0,3,5,6,8,9,10,11,12,13,14,16,17,19,22)$ Your denominators are constant, so you just need to compute the numerators.
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 15:38












  • $begingroup$
    @RossMillikan The goal is primarily to find an equation that can do it.
    $endgroup$
    – user553185
    Dec 1 '18 at 15:40






  • 1




    $begingroup$
    Do you want the denominator to be the least common multiple of 3 and 5? Or is it just there product?
    $endgroup$
    – Fareed AF
    Dec 3 '18 at 11:23




















  • $begingroup$
    Why don't you want to compute them all and then sort them? You can organize the computation us use the fact that subgroups are presorted.
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 15:08










  • $begingroup$
    @RossMillikan Because I'm not writing s computer program
    $endgroup$
    – user553185
    Dec 1 '18 at 15:09










  • $begingroup$
    Then you can just make a list for each $x_1$ going over the values of $x_2$, so you get $(0,3,6,9,12), (5,8,11,14,17), (10,13,16,19,22)$ and pick the smallest value off the front of one, getting $(0,3,5,6,8,9,10,11,12,13,14,16,17,19,22)$ Your denominators are constant, so you just need to compute the numerators.
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 15:38












  • $begingroup$
    @RossMillikan The goal is primarily to find an equation that can do it.
    $endgroup$
    – user553185
    Dec 1 '18 at 15:40






  • 1




    $begingroup$
    Do you want the denominator to be the least common multiple of 3 and 5? Or is it just there product?
    $endgroup$
    – Fareed AF
    Dec 3 '18 at 11:23


















$begingroup$
Why don't you want to compute them all and then sort them? You can organize the computation us use the fact that subgroups are presorted.
$endgroup$
– Ross Millikan
Dec 1 '18 at 15:08




$begingroup$
Why don't you want to compute them all and then sort them? You can organize the computation us use the fact that subgroups are presorted.
$endgroup$
– Ross Millikan
Dec 1 '18 at 15:08












$begingroup$
@RossMillikan Because I'm not writing s computer program
$endgroup$
– user553185
Dec 1 '18 at 15:09




$begingroup$
@RossMillikan Because I'm not writing s computer program
$endgroup$
– user553185
Dec 1 '18 at 15:09












$begingroup$
Then you can just make a list for each $x_1$ going over the values of $x_2$, so you get $(0,3,6,9,12), (5,8,11,14,17), (10,13,16,19,22)$ and pick the smallest value off the front of one, getting $(0,3,5,6,8,9,10,11,12,13,14,16,17,19,22)$ Your denominators are constant, so you just need to compute the numerators.
$endgroup$
– Ross Millikan
Dec 1 '18 at 15:38






$begingroup$
Then you can just make a list for each $x_1$ going over the values of $x_2$, so you get $(0,3,6,9,12), (5,8,11,14,17), (10,13,16,19,22)$ and pick the smallest value off the front of one, getting $(0,3,5,6,8,9,10,11,12,13,14,16,17,19,22)$ Your denominators are constant, so you just need to compute the numerators.
$endgroup$
– Ross Millikan
Dec 1 '18 at 15:38














$begingroup$
@RossMillikan The goal is primarily to find an equation that can do it.
$endgroup$
– user553185
Dec 1 '18 at 15:40




$begingroup$
@RossMillikan The goal is primarily to find an equation that can do it.
$endgroup$
– user553185
Dec 1 '18 at 15:40




1




1




$begingroup$
Do you want the denominator to be the least common multiple of 3 and 5? Or is it just there product?
$endgroup$
– Fareed AF
Dec 3 '18 at 11:23






$begingroup$
Do you want the denominator to be the least common multiple of 3 and 5? Or is it just there product?
$endgroup$
– Fareed AF
Dec 3 '18 at 11:23












1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint:
This isn't an answer it is just a way you can think about it, and I am not sure that it will take you somewhere or not, but name the set $A={0, ..., b_1-1}$
and $B= {0, ..., b_2-1}$



And try taking the sequence $$x_{n,k}=frac{b_1.n+b_2.k}{lcm(b_1,b_2)}$$
Where $n in A$ and $k in B$



Then after that take the $displaystyle min_{n in A,kin B}(x_{n,k})=x_{n_0,k_0}$ and you'll get the first elemnt you want call it $c_1$ (which is $frac{0}{15}$ in your case)



Then exclude this element from the sequence (by excluding $n_0$ from $A$ and $k_0$ from $B$) and take the minimum again and so on...






share|cite|improve this answer











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    1 Answer
    1






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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    Hint:
    This isn't an answer it is just a way you can think about it, and I am not sure that it will take you somewhere or not, but name the set $A={0, ..., b_1-1}$
    and $B= {0, ..., b_2-1}$



    And try taking the sequence $$x_{n,k}=frac{b_1.n+b_2.k}{lcm(b_1,b_2)}$$
    Where $n in A$ and $k in B$



    Then after that take the $displaystyle min_{n in A,kin B}(x_{n,k})=x_{n_0,k_0}$ and you'll get the first elemnt you want call it $c_1$ (which is $frac{0}{15}$ in your case)



    Then exclude this element from the sequence (by excluding $n_0$ from $A$ and $k_0$ from $B$) and take the minimum again and so on...






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Hint:
      This isn't an answer it is just a way you can think about it, and I am not sure that it will take you somewhere or not, but name the set $A={0, ..., b_1-1}$
      and $B= {0, ..., b_2-1}$



      And try taking the sequence $$x_{n,k}=frac{b_1.n+b_2.k}{lcm(b_1,b_2)}$$
      Where $n in A$ and $k in B$



      Then after that take the $displaystyle min_{n in A,kin B}(x_{n,k})=x_{n_0,k_0}$ and you'll get the first elemnt you want call it $c_1$ (which is $frac{0}{15}$ in your case)



      Then exclude this element from the sequence (by excluding $n_0$ from $A$ and $k_0$ from $B$) and take the minimum again and so on...






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint:
        This isn't an answer it is just a way you can think about it, and I am not sure that it will take you somewhere or not, but name the set $A={0, ..., b_1-1}$
        and $B= {0, ..., b_2-1}$



        And try taking the sequence $$x_{n,k}=frac{b_1.n+b_2.k}{lcm(b_1,b_2)}$$
        Where $n in A$ and $k in B$



        Then after that take the $displaystyle min_{n in A,kin B}(x_{n,k})=x_{n_0,k_0}$ and you'll get the first elemnt you want call it $c_1$ (which is $frac{0}{15}$ in your case)



        Then exclude this element from the sequence (by excluding $n_0$ from $A$ and $k_0$ from $B$) and take the minimum again and so on...






        share|cite|improve this answer











        $endgroup$



        Hint:
        This isn't an answer it is just a way you can think about it, and I am not sure that it will take you somewhere or not, but name the set $A={0, ..., b_1-1}$
        and $B= {0, ..., b_2-1}$



        And try taking the sequence $$x_{n,k}=frac{b_1.n+b_2.k}{lcm(b_1,b_2)}$$
        Where $n in A$ and $k in B$



        Then after that take the $displaystyle min_{n in A,kin B}(x_{n,k})=x_{n_0,k_0}$ and you'll get the first elemnt you want call it $c_1$ (which is $frac{0}{15}$ in your case)



        Then exclude this element from the sequence (by excluding $n_0$ from $A$ and $k_0$ from $B$) and take the minimum again and so on...







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 12:31

























        answered Dec 3 '18 at 11:45









        Fareed AFFareed AF

        52612




        52612






























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