I need an opinion and maybe a clarification about my work on the following problem:
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Problem: Suppose that $m_1,m_2,...,m_t$ are positive integers and $a_1,a_2,...,a_t$ are integers.
What condition on $m_1,m_2,...,m_t$ is necessary to guarantee that there is an integer x such that $x ≡ a_i $ mod $m_i$ for all intergers i in the range $1leq ileq r $.
My idea:
I will choose for presentation x=1, $a_1=14$ and $m_1=13$.
Now let k be any positive integer. So that $a_n=a_1+k*(n-1)$ where n>0.
Similarly let $m_n=m_1+k*(n-1)$.
Now for any n, 1 ≡ $a_n $ mod $m_n$.
The solution is that $m_1,m_2,...,m_t$ is pairwise coprime.
In my presentation if I have k=2 then my m's will not be pairwise coprime.
Could you please explain why the answer is pairwise coprime and why will my idea not work.
greatest-common-divisor
asked Dec 27 '18 at 19:41
ValVal
537
$endgroup$
add a comment |
$begingroup$
Problem: Suppose that $m_1,m_2,...,m_t$ are positive integers and $a_1,a_2,...,a_t$ are integers.
What condition on $m_1,m_2,...,m_t$ is necessary to guarantee that there is an integer x such that $x ≡ a_i $ mod $m_i$ for all intergers i in the range $1leq ileq r $.
My idea:
I will choose for presentation x=1, $a_1=14$ and $m_1=13$.
Now let k be any positive integer. So that $a_n=a_1+k*(n-1)$ where n>0.
Similarly let $m_n=m_1+k*(n-1)$.
Now for any n, 1 ≡ $a_n $ mod $m_n$.
The solution is that $m_1,m_2,...,m_t$ is pairwise coprime.
In my presentation if I have k=2 then my m's will not be pairwise coprime.
Could you please explain why the answer is pairwise coprime and why will my idea not work.
greatest-common-divisor
asked Dec 27 '18 at 19:41
ValVal
537
$endgroup$
add a comment |
$begingroup$
Problem: Suppose that $m_1,m_2,...,m_t$ are positive integers and $a_1,a_2,...,a_t$ are integers.
What condition on $m_1,m_2,...,m_t$ is necessary to guarantee that there is an integer x such that $x ≡ a_i $ mod $m_i$ for all intergers i in the range $1leq ileq r $.
My idea:
I will choose for presentation x=1, $a_1=14$ and $m_1=13$.
Now let k be any positive integer. So that $a_n=a_1+k*(n-1)$ where n>0.
Similarly let $m_n=m_1+k*(n-1)$.
Now for any n, 1 ≡ $a_n $ mod $m_n$.
The solution is that $m_1,m_2,...,m_t$ is pairwise coprime.
In my presentation if I have k=2 then my m's will not be pairwise coprime.
Could you please explain why the answer is pairwise coprime and why will my idea not work.
greatest-common-divisor
asked Dec 27 '18 at 19:41
ValVal
537
$endgroup$
Problem: Suppose that $m_1,m_2,...,m_t$ are positive integers and $a_1,a_2,...,a_t$ are integers.
What condition on $m_1,m_2,...,m_t$ is necessary to guarantee that there is an integer x such that $x ≡ a_i $ mod $m_i$ for all intergers i in the range $1leq ileq r $.
My idea:
I will choose for presentation x=1, $a_1=14$ and $m_1=13$.
Now let k be any positive integer. So that $a_n=a_1+k*(n-1)$ where n>0.
Similarly let $m_n=m_1+k*(n-1)$.
Now for any n, 1 ≡ $a_n $ mod $m_n$.
The solution is that $m_1,m_2,...,m_t$ is pairwise coprime.
In my presentation if I have k=2 then my m's will not be pairwise coprime.
Could you please explain why the answer is pairwise coprime and why will my idea not work.
greatest-common-divisor
greatest-common-divisor
asked Dec 27 '18 at 19:41
ValVal
537
asked Dec 27 '18 at 19:41
ValVal
537
edited Dec 27 '18 at 19:46
Val
asked Dec 27 '18 at 19:41
ValVal
537
asked Dec 27 '18 at 19:41
ValVal
537
asked Dec 27 '18 at 19:41
ValVal
537
537
add a comment |
add a comment |
1 Answer
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active
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$begingroup$
Normally you are given the $a$'s and $m$'s and asked to find $x$. If the $m$'s are pairwise coprime, the Chinese Remainder Theorem guarantees a solution. If they are not, there may not be a solution. As an example, take $m_1=4, m_2=6$, which have common factor $2$. Then if $a_1=1,a_2=2$ there is no solution because the first forces $x$ to be odd and the second forces $x$ to be even.
answered Dec 27 '18 at 20:14
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
You said "the first forces x to be even and the second forces x to be odd" is it not the other way around?
$endgroup$
– Val
Dec 28 '18 at 12:15
$begingroup$
@BurLeXyOOnuTz: Yes, a brain glitch. Fixed. Thanks
$endgroup$
– Ross Millikan
Dec 28 '18 at 14:07
add a comment |
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1 Answer
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$begingroup$
Normally you are given the $a$'s and $m$'s and asked to find $x$. If the $m$'s are pairwise coprime, the Chinese Remainder Theorem guarantees a solution. If they are not, there may not be a solution. As an example, take $m_1=4, m_2=6$, which have common factor $2$. Then if $a_1=1,a_2=2$ there is no solution because the first forces $x$ to be odd and the second forces $x$ to be even.
answered Dec 27 '18 at 20:14
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
You said "the first forces x to be even and the second forces x to be odd" is it not the other way around?
$endgroup$
– Val
Dec 28 '18 at 12:15
$begingroup$
@BurLeXyOOnuTz: Yes, a brain glitch. Fixed. Thanks
$endgroup$
– Ross Millikan
Dec 28 '18 at 14:07
add a comment |
$begingroup$
Normally you are given the $a$'s and $m$'s and asked to find $x$. If the $m$'s are pairwise coprime, the Chinese Remainder Theorem guarantees a solution. If they are not, there may not be a solution. As an example, take $m_1=4, m_2=6$, which have common factor $2$. Then if $a_1=1,a_2=2$ there is no solution because the first forces $x$ to be odd and the second forces $x$ to be even.
answered Dec 27 '18 at 20:14
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
You said "the first forces x to be even and the second forces x to be odd" is it not the other way around?
$endgroup$
– Val
Dec 28 '18 at 12:15
$begingroup$
@BurLeXyOOnuTz: Yes, a brain glitch. Fixed. Thanks
$endgroup$
– Ross Millikan
Dec 28 '18 at 14:07
add a comment |
$begingroup$
Normally you are given the $a$'s and $m$'s and asked to find $x$. If the $m$'s are pairwise coprime, the Chinese Remainder Theorem guarantees a solution. If they are not, there may not be a solution. As an example, take $m_1=4, m_2=6$, which have common factor $2$. Then if $a_1=1,a_2=2$ there is no solution because the first forces $x$ to be odd and the second forces $x$ to be even.
answered Dec 27 '18 at 20:14
Ross MillikanRoss Millikan
299k24200374
$endgroup$
Normally you are given the $a$'s and $m$'s and asked to find $x$. If the $m$'s are pairwise coprime, the Chinese Remainder Theorem guarantees a solution. If they are not, there may not be a solution. As an example, take $m_1=4, m_2=6$, which have common factor $2$. Then if $a_1=1,a_2=2$ there is no solution because the first forces $x$ to be odd and the second forces $x$ to be even.
answered Dec 27 '18 at 20:14
Ross MillikanRoss Millikan
299k24200374
edited Dec 28 '18 at 14:07
answered Dec 27 '18 at 20:14
Ross MillikanRoss Millikan
299k24200374
answered Dec 27 '18 at 20:14
Ross MillikanRoss Millikan
299k24200374
answered Dec 27 '18 at 20:14
Ross MillikanRoss Millikan
299k24200374
299k24200374
$begingroup$
You said "the first forces x to be even and the second forces x to be odd" is it not the other way around?
$endgroup$
– Val
Dec 28 '18 at 12:15
$begingroup$
@BurLeXyOOnuTz: Yes, a brain glitch. Fixed. Thanks
$endgroup$
– Ross Millikan
Dec 28 '18 at 14:07
add a comment |
$begingroup$
You said "the first forces x to be even and the second forces x to be odd" is it not the other way around?
$endgroup$
– Val
Dec 28 '18 at 12:15
$begingroup$
@BurLeXyOOnuTz: Yes, a brain glitch. Fixed. Thanks
$endgroup$
– Ross Millikan
Dec 28 '18 at 14:07
$begingroup$
You said "the first forces x to be even and the second forces x to be odd" is it not the other way around?
$endgroup$
– Val
Dec 28 '18 at 12:15
$begingroup$
You said "the first forces x to be even and the second forces x to be odd" is it not the other way around?
$endgroup$
– Val
Dec 28 '18 at 12:15
$begingroup$
@BurLeXyOOnuTz: Yes, a brain glitch. Fixed. Thanks
$endgroup$
– Ross Millikan
Dec 28 '18 at 14:07
$begingroup$
@BurLeXyOOnuTz: Yes, a brain glitch. Fixed. Thanks
$endgroup$
– Ross Millikan
Dec 28 '18 at 14:07
add a comment |
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