How to calculate the sum of the multiples of two fractions in order?
$begingroup$
$frac{a}{b_1}, frac{a}{b_2} = frac{1}{3}, frac{1}{5}$
$x_1 in {0, ..., b_1-1} = {0, 1, 2}$
$x_2 in {0, ..., b_2-1} = {0, 1, 2, 3, 4}$
$$frac{a_1}{b}, frac{a_2}{b} = frac{a_1b_2}{b_1b_2}, frac{a_2b_1}{b_1b_2} = frac{5}{15}, frac{3}{15}$$
$$frac{p}{q} = frac{a_1x_1}{b} + frac{a_2x_2}{b}$$
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
edited Dec 8 '18 at 22:57
Byte Commander
196211
asked Dec 1 '18 at 14:34
user553185user553185
155
$endgroup$
|
show 2 more comments
$begingroup$
$frac{a}{b_1}, frac{a}{b_2} = frac{1}{3}, frac{1}{5}$
$x_1 in {0, ..., b_1-1} = {0, 1, 2}$
$x_2 in {0, ..., b_2-1} = {0, 1, 2, 3, 4}$
$$frac{a_1}{b}, frac{a_2}{b} = frac{a_1b_2}{b_1b_2}, frac{a_2b_1}{b_1b_2} = frac{5}{15}, frac{3}{15}$$
$$frac{p}{q} = frac{a_1x_1}{b} + frac{a_2x_2}{b}$$
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
edited Dec 8 '18 at 22:57
Byte Commander
196211
asked Dec 1 '18 at 14:34
user553185user553185
155
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
$frac{a}{b_1}, frac{a}{b_2} = frac{1}{3}, frac{1}{5}$
$x_1 in {0, ..., b_1-1} = {0, 1, 2}$
$x_2 in {0, ..., b_2-1} = {0, 1, 2, 3, 4}$
$$frac{a_1}{b}, frac{a_2}{b} = frac{a_1b_2}{b_1b_2}, frac{a_2b_1}{b_1b_2} = frac{5}{15}, frac{3}{15}$$
$$frac{p}{q} = frac{a_1x_1}{b} + frac{a_2x_2}{b}$$
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
edited Dec 8 '18 at 22:57
Byte Commander
196211
asked Dec 1 '18 at 14:34
user553185user553185
155
$endgroup$
$frac{a}{b_1}, frac{a}{b_2} = frac{1}{3}, frac{1}{5}$
$x_1 in {0, ..., b_1-1} = {0, 1, 2}$
$x_2 in {0, ..., b_2-1} = {0, 1, 2, 3, 4}$
$$frac{a_1}{b}, frac{a_2}{b} = frac{a_1b_2}{b_1b_2}, frac{a_2b_1}{b_1b_2} = frac{5}{15}, frac{3}{15}$$
$$frac{p}{q} = frac{a_1x_1}{b} + frac{a_2x_2}{b}$$
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
edited Dec 8 '18 at 22:57
Byte Commander
196211
asked Dec 1 '18 at 14:34
user553185user553185
155
edited Dec 8 '18 at 22:57
Byte Commander
196211
asked Dec 1 '18 at 14:34
user553185user553185
155
edited Dec 8 '18 at 22:57
Byte Commander
196211
edited Dec 8 '18 at 22:57
Byte Commander
196211
edited Dec 8 '18 at 22:57
Byte Commander
196211
196211
asked Dec 1 '18 at 14:34
user553185user553185
155
asked Dec 1 '18 at 14:34
user553185user553185
155
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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...
answered Dec 3 '18 at 11:45
Fareed AFFareed AF
52612
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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...
answered Dec 3 '18 at 11:45
Fareed AFFareed AF
52612
$endgroup$
add a comment |
$begingroup$
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...
answered Dec 3 '18 at 11:45
Fareed AFFareed AF
52612
$endgroup$
add a comment |
$begingroup$
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...
answered Dec 3 '18 at 11:45
Fareed AFFareed AF
52612
$endgroup$
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...
answered Dec 3 '18 at 11:45
Fareed AFFareed AF
52612
edited Dec 3 '18 at 12:31
answered Dec 3 '18 at 11:45
Fareed AFFareed AF
52612
answered Dec 3 '18 at 11:45
Fareed AFFareed AF
52612
answered Dec 3 '18 at 11:45
Fareed AFFareed AF
52612
52612
add a comment |
add a comment |
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$begingroup$
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