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I am trying to solve the following equation:

$(a*n + c) \mod (b-n) \equiv 0$

and $n$ must be the lowest value in $[0, b-1]$

for example $a=17$, $c=-59$ and $b=128$, the solution is $n=55$

$n=b-1$ will be always a solution, because $m \mod 1 \equiv 0$

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Equivalently, you want to solve in integers $$ y (b-n) = an+c$$ This is equivalent to $$ (y+a)(b-n) = ab+c $$ You want $b-n$ to be as large as possible subject to $b-n \le b$. Thus you want to factor $ab+c$ and take its largest divisor $\le b$. In your example, $ab+c = 2117 = 29 \cdot 73$ whose largest divisor $\le 128$ is $73$, so $n = 128 - 73 = 55$.

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