Solve the congruence system $ xequiv m_i-1 pmod{m_i},$ for $,i = 1,ldots, k$
Find natural number $x$ so that
$$xequiv 9pmod{10},quad xequiv8pmod9,quad ...,quad xequiv 1pmod2$$
elementary-number-theory
edited Sep 4 '17 at 21:18
Bill Dubuque
208k29190626
asked Mar 16 '14 at 20:37
user119081
373
add a comment |
Find natural number $x$ so that
$$xequiv 9pmod{10},quad xequiv8pmod9,quad ...,quad xequiv 1pmod2$$
elementary-number-theory
edited Sep 4 '17 at 21:18
Bill Dubuque
208k29190626
asked Mar 16 '14 at 20:37
user119081
373
add a comment |
Find natural number $x$ so that
$$xequiv 9pmod{10},quad xequiv8pmod9,quad ...,quad xequiv 1pmod2$$
elementary-number-theory
edited Sep 4 '17 at 21:18
Bill Dubuque
208k29190626
asked Mar 16 '14 at 20:37
user119081
373
Find natural number $x$ so that
$$xequiv 9pmod{10},quad xequiv8pmod9,quad ...,quad xequiv 1pmod2$$
elementary-number-theory
elementary-number-theory
edited Sep 4 '17 at 21:18
Bill Dubuque
208k29190626
asked Mar 16 '14 at 20:37
user119081
373
edited Sep 4 '17 at 21:18
Bill Dubuque
208k29190626
asked Mar 16 '14 at 20:37
user119081
373
edited Sep 4 '17 at 21:18
Bill Dubuque
208k29190626
edited Sep 4 '17 at 21:18
Bill Dubuque
208k29190626
edited Sep 4 '17 at 21:18
Bill Dubuque
208k29190626
208k29190626
asked Mar 16 '14 at 20:37
user119081
373
asked Mar 16 '14 at 20:37
user119081
373
asked Mar 16 '14 at 20:37
user119081
373
373
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Hint: The unnatural number $-1$ works.
answered Mar 16 '14 at 20:41
André Nicolas
451k36421805
I ask for natural number not integer
– user119081
Mar 16 '14 at 20:44
@user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
– Charles
Mar 16 '14 at 20:45
Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
– André Nicolas
Mar 16 '14 at 20:46
I believe this should be exist a natural solution by extended Euclidean algorithm
– user119081
Mar 16 '14 at 20:48
1
While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
– André Nicolas
Mar 16 '14 at 20:52
|
show 3 more comments
Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT
$$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
end{align}qquadqquad$$
The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)
answered Mar 16 '14 at 20:52
Bill Dubuque
208k29190626
It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
– Bill Dubuque
Mar 16 '14 at 21:11
add a comment |
begin{align}
x &equiv 9 pmod{10} \
x &equiv 8 pmod 9 \
x &equiv 7 pmod 8 \
x &equiv 6 pmod 7 \
x &equiv 5 pmod 6 \
x &equiv 4 pmod 5 \
x &equiv 3 pmod 4 \
x &equiv 2 pmod 3 \
x &equiv 1 pmod 2 \
end{align}
Is equivalent to
begin{align}
x &equiv -1 pmod{10} \
x &equiv -1 pmod 9 \
x &equiv -1 pmod 8 \
x &equiv -1 pmod 7 \
x &equiv -1 pmod 6 \
x &equiv -1 pmod 5 \
x &equiv -1 pmod 4 \
x &equiv -1 pmod 3 \
x &equiv -1 pmod 2 \
end{align}
which is equivalent to
$$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$
$$x equiv -1 pmod{2520}$$
$$x equiv 2519 pmod{2520}$$
answered Feb 16 '17 at 14:40
steven gregory
17.7k32257
add a comment |
Your Answer
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3 Answers
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Hint: The unnatural number $-1$ works.
answered Mar 16 '14 at 20:41
André Nicolas
451k36421805
I ask for natural number not integer
– user119081
Mar 16 '14 at 20:44
@user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
– Charles
Mar 16 '14 at 20:45
Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
– André Nicolas
Mar 16 '14 at 20:46
I believe this should be exist a natural solution by extended Euclidean algorithm
– user119081
Mar 16 '14 at 20:48
1
While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
– André Nicolas
Mar 16 '14 at 20:52
|
show 3 more comments
Hint: The unnatural number $-1$ works.
answered Mar 16 '14 at 20:41
André Nicolas
451k36421805
I ask for natural number not integer
– user119081
Mar 16 '14 at 20:44
@user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
– Charles
Mar 16 '14 at 20:45
Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
– André Nicolas
Mar 16 '14 at 20:46
I believe this should be exist a natural solution by extended Euclidean algorithm
– user119081
Mar 16 '14 at 20:48
1
While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
– André Nicolas
Mar 16 '14 at 20:52
|
show 3 more comments
Hint: The unnatural number $-1$ works.
answered Mar 16 '14 at 20:41
André Nicolas
451k36421805
Hint: The unnatural number $-1$ works.
answered Mar 16 '14 at 20:41
André Nicolas
451k36421805
answered Mar 16 '14 at 20:41
André Nicolas
451k36421805
answered Mar 16 '14 at 20:41
André Nicolas
451k36421805
answered Mar 16 '14 at 20:41
André Nicolas
451k36421805
451k36421805
I ask for natural number not integer
– user119081
Mar 16 '14 at 20:44
@user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
– Charles
Mar 16 '14 at 20:45
Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
– André Nicolas
Mar 16 '14 at 20:46
I believe this should be exist a natural solution by extended Euclidean algorithm
– user119081
Mar 16 '14 at 20:48
1
While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
– André Nicolas
Mar 16 '14 at 20:52
|
show 3 more comments
I ask for natural number not integer
– user119081
Mar 16 '14 at 20:44
@user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
– Charles
Mar 16 '14 at 20:45
Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
– André Nicolas
Mar 16 '14 at 20:46
I believe this should be exist a natural solution by extended Euclidean algorithm
– user119081
Mar 16 '14 at 20:48
1
While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
– André Nicolas
Mar 16 '14 at 20:52
I ask for natural number not integer
– user119081
Mar 16 '14 at 20:44
I ask for natural number not integer
– user119081
Mar 16 '14 at 20:44
@user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
– Charles
Mar 16 '14 at 20:45
@user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
– Charles
Mar 16 '14 at 20:45
Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
– André Nicolas
Mar 16 '14 at 20:46
Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
– André Nicolas
Mar 16 '14 at 20:46
I believe this should be exist a natural solution by extended Euclidean algorithm
– user119081
Mar 16 '14 at 20:48
I believe this should be exist a natural solution by extended Euclidean algorithm
– user119081
Mar 16 '14 at 20:48
1
1
While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
– André Nicolas
Mar 16 '14 at 20:52
While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
– André Nicolas
Mar 16 '14 at 20:52
|
show 3 more comments
Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT
$$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
end{align}qquadqquad$$
The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)
answered Mar 16 '14 at 20:52
Bill Dubuque
208k29190626
It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
– Bill Dubuque
Mar 16 '14 at 21:11
add a comment |
Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT
$$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
end{align}qquadqquad$$
The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)
answered Mar 16 '14 at 20:52
Bill Dubuque
208k29190626
It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
– Bill Dubuque
Mar 16 '14 at 21:11
add a comment |
Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT
$$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
end{align}qquadqquad$$
The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)
answered Mar 16 '14 at 20:52
Bill Dubuque
208k29190626
Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT
$$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
end{align}qquadqquad$$
The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)
answered Mar 16 '14 at 20:52
Bill Dubuque
208k29190626
edited Sep 4 '17 at 21:20
answered Mar 16 '14 at 20:52
Bill Dubuque
208k29190626
answered Mar 16 '14 at 20:52
Bill Dubuque
208k29190626
answered Mar 16 '14 at 20:52
Bill Dubuque
208k29190626
208k29190626
It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
– Bill Dubuque
Mar 16 '14 at 21:11
add a comment |
It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
– Bill Dubuque
Mar 16 '14 at 21:11
It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
– Bill Dubuque
Mar 16 '14 at 21:11
It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
– Bill Dubuque
Mar 16 '14 at 21:11
add a comment |
begin{align}
x &equiv 9 pmod{10} \
x &equiv 8 pmod 9 \
x &equiv 7 pmod 8 \
x &equiv 6 pmod 7 \
x &equiv 5 pmod 6 \
x &equiv 4 pmod 5 \
x &equiv 3 pmod 4 \
x &equiv 2 pmod 3 \
x &equiv 1 pmod 2 \
end{align}
Is equivalent to
begin{align}
x &equiv -1 pmod{10} \
x &equiv -1 pmod 9 \
x &equiv -1 pmod 8 \
x &equiv -1 pmod 7 \
x &equiv -1 pmod 6 \
x &equiv -1 pmod 5 \
x &equiv -1 pmod 4 \
x &equiv -1 pmod 3 \
x &equiv -1 pmod 2 \
end{align}
which is equivalent to
$$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$
$$x equiv -1 pmod{2520}$$
$$x equiv 2519 pmod{2520}$$
answered Feb 16 '17 at 14:40
steven gregory
17.7k32257
add a comment |
begin{align}
x &equiv 9 pmod{10} \
x &equiv 8 pmod 9 \
x &equiv 7 pmod 8 \
x &equiv 6 pmod 7 \
x &equiv 5 pmod 6 \
x &equiv 4 pmod 5 \
x &equiv 3 pmod 4 \
x &equiv 2 pmod 3 \
x &equiv 1 pmod 2 \
end{align}
Is equivalent to
begin{align}
x &equiv -1 pmod{10} \
x &equiv -1 pmod 9 \
x &equiv -1 pmod 8 \
x &equiv -1 pmod 7 \
x &equiv -1 pmod 6 \
x &equiv -1 pmod 5 \
x &equiv -1 pmod 4 \
x &equiv -1 pmod 3 \
x &equiv -1 pmod 2 \
end{align}
which is equivalent to
$$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$
$$x equiv -1 pmod{2520}$$
$$x equiv 2519 pmod{2520}$$
answered Feb 16 '17 at 14:40
steven gregory
17.7k32257
add a comment |
begin{align}
x &equiv 9 pmod{10} \
x &equiv 8 pmod 9 \
x &equiv 7 pmod 8 \
x &equiv 6 pmod 7 \
x &equiv 5 pmod 6 \
x &equiv 4 pmod 5 \
x &equiv 3 pmod 4 \
x &equiv 2 pmod 3 \
x &equiv 1 pmod 2 \
end{align}
Is equivalent to
begin{align}
x &equiv -1 pmod{10} \
x &equiv -1 pmod 9 \
x &equiv -1 pmod 8 \
x &equiv -1 pmod 7 \
x &equiv -1 pmod 6 \
x &equiv -1 pmod 5 \
x &equiv -1 pmod 4 \
x &equiv -1 pmod 3 \
x &equiv -1 pmod 2 \
end{align}
which is equivalent to
$$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$
$$x equiv -1 pmod{2520}$$
$$x equiv 2519 pmod{2520}$$
answered Feb 16 '17 at 14:40
steven gregory
17.7k32257
begin{align}
x &equiv 9 pmod{10} \
x &equiv 8 pmod 9 \
x &equiv 7 pmod 8 \
x &equiv 6 pmod 7 \
x &equiv 5 pmod 6 \
x &equiv 4 pmod 5 \
x &equiv 3 pmod 4 \
x &equiv 2 pmod 3 \
x &equiv 1 pmod 2 \
end{align}
Is equivalent to
begin{align}
x &equiv -1 pmod{10} \
x &equiv -1 pmod 9 \
x &equiv -1 pmod 8 \
x &equiv -1 pmod 7 \
x &equiv -1 pmod 6 \
x &equiv -1 pmod 5 \
x &equiv -1 pmod 4 \
x &equiv -1 pmod 3 \
x &equiv -1 pmod 2 \
end{align}
which is equivalent to
$$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$
$$x equiv -1 pmod{2520}$$
$$x equiv 2519 pmod{2520}$$
answered Feb 16 '17 at 14:40
steven gregory
17.7k32257
edited Feb 17 '17 at 7:58
answered Feb 16 '17 at 14:40
steven gregory
17.7k32257
answered Feb 16 '17 at 14:40
steven gregory
17.7k32257
answered Feb 16 '17 at 14:40
steven gregory
17.7k32257
17.7k32257
add a comment |
add a comment |
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