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This question is created taking reference from this question. I am aware of Chinese remainder theorem and used that to solve problems of this type. However, answer given in the site referred uses a special property and I want to see how it is derived. Therefore, posting it as question here. I am rewriting the question and answer in a general form and I hope i am writing in the right way (note that my knowledge on modular arithmetic is limited).

Consider

$N \equiv r_1 \pmod {d_1}\\N \equiv r_2 \pmod {d_2}\\N \equiv r_3 \pmod {d_3}\\N \equiv r_4 \pmod {d_4}\\$

So, to find N, we can use Chinese remainder theorem. But, suppose

$d_1-r_1=d_2-r_2=d_3-r_3=d_4-r_4=k$

So, based on the answer provided in the referred site, the required number can be written as

$N = m \times LCM(d_1,d_2,d_3,d_4)-k$ where m is a positive integer

How this can be proven mathematically?

My attempt
I have rewritten it in the following form and was trying to solve.

$N=n_1d_1+r_1\\N=n_2d_2+r_2\\N=n_3d_3+r_3\\N=n_4d_4+r_4$

Let $LCM(d_1,d_2,d_3,d_4)=L$
Then

$N=L+r_1\\N=L+r_2\\N=L+r_3\\N=L+r_4$

please help.

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    You already know that $d_i \mid LCM(d_1,d_2,d_3,d_4)$, so what is $N\; mod\; d_i$? – 2017-01-22
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    By $N\text{ mod }d_i$ , you mean the remainder of $N/d_i$? – 2017-01-22

1 Answers 1

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$\, N\!+\!k\equiv 0\pmod{\!d_i}\iff d_i\mid N\!+\!k\iff {\rm lcm}\{d_i\}\mid N\!+\!k\iff N\equiv -k\pmod{{\rm lcm}\{d_i\}}$

Or, equivalently: $\,\ N\equiv -k\pmod{d_i}\iff N\equiv -k\pmod{{\rm lcm}\{d_i\}},\, $ which expresses it as a ubiquitous constant-case optimization of the Chinese Remainder Theorem (CRT).

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    the symbol, $d_i|N+k$ means, $d_i$ is divisible by $(N+k)$, right? – 2017-01-24
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    @Kiran The reverse. $\ a\mid b\,$ means $\,a\,$ divides $\,b,\,$ i.e. $\, n a = b\,$ for some integer $\,n.\,$ This is standard notation in number theory. You can find many examples in the linked posts. – 2017-01-24
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    ok, thanks for clarifying. – 2017-01-24
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    your answer is really helpful. thanks. i am now learning more about number theory through this website. – 2017-01-24
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    @Kiran Generally if we can arithmetically "tweak" the $r_i$ so they all become the same then we have found a solution. Above we tweaked them by negating them all and noticing they all are the same, i.e. $\, -N\equiv -r_i\equiv k\pmod{d_i}.\,$ – 2017-01-24