<s>[INST] On Chinese remainder theorem Suppose we know $x\equiv 3 \bmod 11$ and $x\equiv 7\bmod 13$ and $0<x<143$ holds then CRT gives that  $x=3\times 13[13^{-1}\bmod 11] + 7\times 11[11^{-1}\bmod 13]=39\times 6 + 77\times 6=696$ gives a solution but it is not within $0$ and $143$. So we take $696\bmod 143$ and choose $124$ as solution.
My query is say instead of $3$ we choose $3+11k$ and instead of $7$ we choose $7+13\ell$ for some $k,\ell\in\Bbb Z$ can we still recover $124$? 
Instead of $696$ is there a direct way to get $124$?
We have $696=124+4(143)$. In general what is the quantity that goes in instead of $4$?
 [/INST] The direct way to $124$ would be to start with $7$ and see how many $13$s to add to make it into a number that is $3\bmod 11$. 
Since $13\equiv \color{red}2 \bmod 11$, we can see that we need $(3-7)/\color{red}2 = -2\equiv 9\bmod 11$ copies of $13$. $7+9\cdot 13 = 7+117 = 124$.
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