having problem in congruence
if $m= k_1\cdot k_2$
then $a\equiv b\bmod (k_1)$
$a\equiv b_1\bmod (k_2)$
the can i write that ..
$a\equiv b\cdot b_1\bmod (m)$
i am confused ??!
then $a\equiv b\bmod (k_1)$
$a\equiv b_1\bmod (k_2)$
the can i write that ..
$a\equiv b\cdot b_1\bmod (m)$
i am confused ??!
Re: having problem in congruence
Its not true. For example, take $a=8, k_1=2, k_2=3$
Every logical solution to a problem has its own beauty.
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Re: having problem in congruence
is there any property of congruence close to this ?!
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Re: having problem in congruence
two properties are like these...
$1.$ if $m=k_1\cdot k_2$ and $gcd\left(k_1, k_2\right)=1$...
$a\equiv b\bmod(k_1)$
$a\equiv b\bmod(k_2)$
then $a\equiv b\bmod(m)$
$2.$ $a\equiv b\bmod(m)$
$c\equiv d\bmod(m)$
then $ac\equiv bd\bmod(m)$
$1.$ if $m=k_1\cdot k_2$ and $gcd\left(k_1, k_2\right)=1$...
$a\equiv b\bmod(k_1)$
$a\equiv b\bmod(k_2)$
then $a\equiv b\bmod(m)$
$2.$ $a\equiv b\bmod(m)$
$c\equiv d\bmod(m)$
then $ac\equiv bd\bmod(m)$
Last edited by HandaramTheGreat on Sun Dec 26, 2010 4:14 pm, edited 1 time in total.
Re: having problem in congruence
$ m $ doesnt have to be the lcm of $k_1$ and $k_2 $
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Re: having problem in congruence
this is only for first property...HandaramTheGreat wrote:if $m=k_1\cdot k_2$ and $gcd\left(k_1, k_2\right)=1$...
Re: having problem in congruence
i know it was for the 1st property . so $m$ has to be the lcm of $k_1$ and $k_2$HandaramTheGreat wrote:HandaramTheGreat wrote:if $m=k_1\cdot k_2$ and $gcd\left(k_1, k_2\right)=1$...
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- Posts:135
- Joined:Thu Dec 09, 2010 12:10 pm
Re: having problem in congruence
then what's it? can't understand what you want to say...m doesnt have to be the lcm of k1 and k2
Re: having problem in congruence
this comment is right .....?tushar7 wrote:i know it was for the 1st property . so $m$ has to be the lcm of $k_1$ and $k_2$HandaramTheGreat wrote:HandaramTheGreat wrote:if $m=k_1\cdot k_2$ and $gcd\left(k_1, k_2\right)=1$...
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- Posts:135
- Joined:Thu Dec 09, 2010 12:10 pm
Re: having problem in congruence
hm... if $k_1$ and $k_2$ is coprime then lcm of $k_1$ and $k_2$ is their product, isn't it?