Solve linear congruences
$begingroup$
Solve linear congruences system
- $11x equiv 10 mod 12$
- $14x equiv 10 mod 15$
- $20x equiv 10 mod 21$
We need to find x that is closest to 1200. The correct solution is 1250.
This is the way I have done it:
First I converted all equations to their correct shape:
- $x equiv 2 mod 12$
- $x equiv 5 mod 15$
- $x equiv 11 mod 21$
Then I do $x = 2 + 12*k$, which I put into the second equation and I get $k = 4 mod 15$. I put that back into first one and I get $x = 5 + 18*h$. I put that into the third one and I get $h = 6 + 7*m$. But how am I supposed to solve this. I also know the solution needs to be in modulo $420$, because $lcm(12,15,21)=420$.
discrete-mathematics modular-arithmetic congruence-relations
edited Dec 16 '18 at 10:30
gimusi
92.9k84494
asked Dec 16 '18 at 8:43
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
Solve linear congruences system
- $11x equiv 10 mod 12$
- $14x equiv 10 mod 15$
- $20x equiv 10 mod 21$
We need to find x that is closest to 1200. The correct solution is 1250.
This is the way I have done it:
First I converted all equations to their correct shape:
- $x equiv 2 mod 12$
- $x equiv 5 mod 15$
- $x equiv 11 mod 21$
Then I do $x = 2 + 12*k$, which I put into the second equation and I get $k = 4 mod 15$. I put that back into first one and I get $x = 5 + 18*h$. I put that into the third one and I get $h = 6 + 7*m$. But how am I supposed to solve this. I also know the solution needs to be in modulo $420$, because $lcm(12,15,21)=420$.
discrete-mathematics modular-arithmetic congruence-relations
edited Dec 16 '18 at 10:30
gimusi
92.9k84494
asked Dec 16 '18 at 8:43
ponikoliponikoli
416
$endgroup$
$begingroup$
To be sure that the system has solution (12,15,21) should be coprimes. To apply CRT we need to write at first the equivalent system for the factors $12=3cdot 4$, $15=3cdot 5$ and $21=3cdot 7$ and check that it is consistent.
$endgroup$
– gimusi
Dec 16 '18 at 11:34
add a comment |
$begingroup$
Solve linear congruences system
- $11x equiv 10 mod 12$
- $14x equiv 10 mod 15$
- $20x equiv 10 mod 21$
We need to find x that is closest to 1200. The correct solution is 1250.
This is the way I have done it:
First I converted all equations to their correct shape:
- $x equiv 2 mod 12$
- $x equiv 5 mod 15$
- $x equiv 11 mod 21$
Then I do $x = 2 + 12*k$, which I put into the second equation and I get $k = 4 mod 15$. I put that back into first one and I get $x = 5 + 18*h$. I put that into the third one and I get $h = 6 + 7*m$. But how am I supposed to solve this. I also know the solution needs to be in modulo $420$, because $lcm(12,15,21)=420$.
discrete-mathematics modular-arithmetic congruence-relations
edited Dec 16 '18 at 10:30
gimusi
92.9k84494
asked Dec 16 '18 at 8:43
ponikoliponikoli
416
$endgroup$
Solve linear congruences system
- $11x equiv 10 mod 12$
- $14x equiv 10 mod 15$
- $20x equiv 10 mod 21$
We need to find x that is closest to 1200. The correct solution is 1250.
This is the way I have done it:
First I converted all equations to their correct shape:
- $x equiv 2 mod 12$
- $x equiv 5 mod 15$
- $x equiv 11 mod 21$
Then I do $x = 2 + 12*k$, which I put into the second equation and I get $k = 4 mod 15$. I put that back into first one and I get $x = 5 + 18*h$. I put that into the third one and I get $h = 6 + 7*m$. But how am I supposed to solve this. I also know the solution needs to be in modulo $420$, because $lcm(12,15,21)=420$.
discrete-mathematics modular-arithmetic congruence-relations
discrete-mathematics modular-arithmetic congruence-relations
edited Dec 16 '18 at 10:30
gimusi
92.9k84494
asked Dec 16 '18 at 8:43
ponikoliponikoli
416
edited Dec 16 '18 at 10:30
gimusi
92.9k84494
asked Dec 16 '18 at 8:43
ponikoliponikoli
416
edited Dec 16 '18 at 10:30
gimusi
92.9k84494
edited Dec 16 '18 at 10:30
gimusi
92.9k84494
edited Dec 16 '18 at 10:30
gimusi
92.9k84494
92.9k84494
asked Dec 16 '18 at 8:43
ponikoliponikoli
416
asked Dec 16 '18 at 8:43
ponikoliponikoli
416
asked Dec 16 '18 at 8:43
ponikoliponikoli
416
416
$begingroup$
To be sure that the system has solution (12,15,21) should be coprimes. To apply CRT we need to write at first the equivalent system for the factors $12=3cdot 4$, $15=3cdot 5$ and $21=3cdot 7$ and check that it is consistent.
$endgroup$
– gimusi
Dec 16 '18 at 11:34
add a comment |
$begingroup$
To be sure that the system has solution (12,15,21) should be coprimes. To apply CRT we need to write at first the equivalent system for the factors $12=3cdot 4$, $15=3cdot 5$ and $21=3cdot 7$ and check that it is consistent.
$endgroup$
– gimusi
Dec 16 '18 at 11:34
$begingroup$
To be sure that the system has solution (12,15,21) should be coprimes. To apply CRT we need to write at first the equivalent system for the factors $12=3cdot 4$, $15=3cdot 5$ and $21=3cdot 7$ and check that it is consistent.
$endgroup$
– gimusi
Dec 16 '18 at 11:34
$begingroup$
To be sure that the system has solution (12,15,21) should be coprimes. To apply CRT we need to write at first the equivalent system for the factors $12=3cdot 4$, $15=3cdot 5$ and $21=3cdot 7$ and check that it is consistent.
$endgroup$
– gimusi
Dec 16 '18 at 11:34
add a comment |
2 Answers
2
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oldest
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$begingroup$
You can plug $x=12a+2$ into the second and third equations:
$$begin{cases}12a+2equiv 5 pmod{15} \ 12a+2equiv 11 pmod{21}end{cases} Rightarrow begin{cases}aequiv 4 pmod{5} \ aequiv 6 pmod{7}end{cases}$$
Now plug $a=5b+4$ into the last equation:
$$5b+4equiv 6 pmod{7} Rightarrow bequiv 6 pmod{7} Rightarrow b=7c+6$$
Now go back:
$$a=5b+4=5(7c+6)+4=35c+34;\
x=12a+2=12(35c+34)+34=420c+410 Rightarrow \
xequiv 410 pmod{420}.$$
Verifying $x=410$:
$$11cdot 410equiv 4510 equiv 375cdot 12+10equiv 10 pmod{12};\
14cdot 410equiv 5740 equiv 382cdot 15+10equiv 10 pmod{15};\
20cdot 410equiv 8200 equiv 390cdot 21+10equiv 10 pmod{21}.$$
You can check $x=420cdot 2+410=1250$ youserlf.
answered Dec 16 '18 at 11:35
farruhotafarruhota
20.7k2740
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$begingroup$
Hint $ {-}xequiv 10pmod {!a,b,c}iff a,b,cmid x!+!10iff{rm lcm}(a,b,c)mid x!+!10$
${rm lcm}(12,15,21) = 3, {rm lcm}(4,5,7) = 3cdot 4cdot 5cdot 7 = 20cdot 21 = 420,, $ hence $,xequiv -10pmod{!420}$
answered Dec 16 '18 at 21:09
Bill DubuqueBill Dubuque
212k29195650
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2 Answers
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$begingroup$
You can plug $x=12a+2$ into the second and third equations:
$$begin{cases}12a+2equiv 5 pmod{15} \ 12a+2equiv 11 pmod{21}end{cases} Rightarrow begin{cases}aequiv 4 pmod{5} \ aequiv 6 pmod{7}end{cases}$$
Now plug $a=5b+4$ into the last equation:
$$5b+4equiv 6 pmod{7} Rightarrow bequiv 6 pmod{7} Rightarrow b=7c+6$$
Now go back:
$$a=5b+4=5(7c+6)+4=35c+34;\
x=12a+2=12(35c+34)+34=420c+410 Rightarrow \
xequiv 410 pmod{420}.$$
Verifying $x=410$:
$$11cdot 410equiv 4510 equiv 375cdot 12+10equiv 10 pmod{12};\
14cdot 410equiv 5740 equiv 382cdot 15+10equiv 10 pmod{15};\
20cdot 410equiv 8200 equiv 390cdot 21+10equiv 10 pmod{21}.$$
You can check $x=420cdot 2+410=1250$ youserlf.
answered Dec 16 '18 at 11:35
farruhotafarruhota
20.7k2740
$endgroup$
add a comment |
$begingroup$
You can plug $x=12a+2$ into the second and third equations:
$$begin{cases}12a+2equiv 5 pmod{15} \ 12a+2equiv 11 pmod{21}end{cases} Rightarrow begin{cases}aequiv 4 pmod{5} \ aequiv 6 pmod{7}end{cases}$$
Now plug $a=5b+4$ into the last equation:
$$5b+4equiv 6 pmod{7} Rightarrow bequiv 6 pmod{7} Rightarrow b=7c+6$$
Now go back:
$$a=5b+4=5(7c+6)+4=35c+34;\
x=12a+2=12(35c+34)+34=420c+410 Rightarrow \
xequiv 410 pmod{420}.$$
Verifying $x=410$:
$$11cdot 410equiv 4510 equiv 375cdot 12+10equiv 10 pmod{12};\
14cdot 410equiv 5740 equiv 382cdot 15+10equiv 10 pmod{15};\
20cdot 410equiv 8200 equiv 390cdot 21+10equiv 10 pmod{21}.$$
You can check $x=420cdot 2+410=1250$ youserlf.
answered Dec 16 '18 at 11:35
farruhotafarruhota
20.7k2740
$endgroup$
add a comment |
$begingroup$
You can plug $x=12a+2$ into the second and third equations:
$$begin{cases}12a+2equiv 5 pmod{15} \ 12a+2equiv 11 pmod{21}end{cases} Rightarrow begin{cases}aequiv 4 pmod{5} \ aequiv 6 pmod{7}end{cases}$$
Now plug $a=5b+4$ into the last equation:
$$5b+4equiv 6 pmod{7} Rightarrow bequiv 6 pmod{7} Rightarrow b=7c+6$$
Now go back:
$$a=5b+4=5(7c+6)+4=35c+34;\
x=12a+2=12(35c+34)+34=420c+410 Rightarrow \
xequiv 410 pmod{420}.$$
Verifying $x=410$:
$$11cdot 410equiv 4510 equiv 375cdot 12+10equiv 10 pmod{12};\
14cdot 410equiv 5740 equiv 382cdot 15+10equiv 10 pmod{15};\
20cdot 410equiv 8200 equiv 390cdot 21+10equiv 10 pmod{21}.$$
You can check $x=420cdot 2+410=1250$ youserlf.
answered Dec 16 '18 at 11:35
farruhotafarruhota
20.7k2740
$endgroup$
You can plug $x=12a+2$ into the second and third equations:
$$begin{cases}12a+2equiv 5 pmod{15} \ 12a+2equiv 11 pmod{21}end{cases} Rightarrow begin{cases}aequiv 4 pmod{5} \ aequiv 6 pmod{7}end{cases}$$
Now plug $a=5b+4$ into the last equation:
$$5b+4equiv 6 pmod{7} Rightarrow bequiv 6 pmod{7} Rightarrow b=7c+6$$
Now go back:
$$a=5b+4=5(7c+6)+4=35c+34;\
x=12a+2=12(35c+34)+34=420c+410 Rightarrow \
xequiv 410 pmod{420}.$$
Verifying $x=410$:
$$11cdot 410equiv 4510 equiv 375cdot 12+10equiv 10 pmod{12};\
14cdot 410equiv 5740 equiv 382cdot 15+10equiv 10 pmod{15};\
20cdot 410equiv 8200 equiv 390cdot 21+10equiv 10 pmod{21}.$$
You can check $x=420cdot 2+410=1250$ youserlf.
answered Dec 16 '18 at 11:35
farruhotafarruhota
20.7k2740
answered Dec 16 '18 at 11:35
farruhotafarruhota
20.7k2740
answered Dec 16 '18 at 11:35
farruhotafarruhota
20.7k2740
answered Dec 16 '18 at 11:35
farruhotafarruhota
20.7k2740
20.7k2740
add a comment |
add a comment |
$begingroup$
Hint $ {-}xequiv 10pmod {!a,b,c}iff a,b,cmid x!+!10iff{rm lcm}(a,b,c)mid x!+!10$
${rm lcm}(12,15,21) = 3, {rm lcm}(4,5,7) = 3cdot 4cdot 5cdot 7 = 20cdot 21 = 420,, $ hence $,xequiv -10pmod{!420}$
answered Dec 16 '18 at 21:09
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
add a comment |
$begingroup$
Hint $ {-}xequiv 10pmod {!a,b,c}iff a,b,cmid x!+!10iff{rm lcm}(a,b,c)mid x!+!10$
${rm lcm}(12,15,21) = 3, {rm lcm}(4,5,7) = 3cdot 4cdot 5cdot 7 = 20cdot 21 = 420,, $ hence $,xequiv -10pmod{!420}$
answered Dec 16 '18 at 21:09
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
add a comment |
$begingroup$
Hint $ {-}xequiv 10pmod {!a,b,c}iff a,b,cmid x!+!10iff{rm lcm}(a,b,c)mid x!+!10$
${rm lcm}(12,15,21) = 3, {rm lcm}(4,5,7) = 3cdot 4cdot 5cdot 7 = 20cdot 21 = 420,, $ hence $,xequiv -10pmod{!420}$
answered Dec 16 '18 at 21:09
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
Hint $ {-}xequiv 10pmod {!a,b,c}iff a,b,cmid x!+!10iff{rm lcm}(a,b,c)mid x!+!10$
${rm lcm}(12,15,21) = 3, {rm lcm}(4,5,7) = 3cdot 4cdot 5cdot 7 = 20cdot 21 = 420,, $ hence $,xequiv -10pmod{!420}$
answered Dec 16 '18 at 21:09
Bill DubuqueBill Dubuque
212k29195650
edited Dec 16 '18 at 21:32
answered Dec 16 '18 at 21:09
Bill DubuqueBill Dubuque
212k29195650
answered Dec 16 '18 at 21:09
Bill DubuqueBill Dubuque
212k29195650
answered Dec 16 '18 at 21:09
Bill DubuqueBill Dubuque
212k29195650
212k29195650
add a comment |
add a comment |
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$begingroup$
To be sure that the system has solution (12,15,21) should be coprimes. To apply CRT we need to write at first the equivalent system for the factors $12=3cdot 4$, $15=3cdot 5$ and $21=3cdot 7$ and check that it is consistent.
$endgroup$
– gimusi
Dec 16 '18 at 11:34