Modular Ratio

$83 \equiv 75n \pmod{32}$

$19 \equiv 11n \pmod{32}$

Then $19 = 11n + 32m$ for some integers $n,m$ .

By using Euclidean Algorithm , we can solve this linear diophantine equation:

$32 = 11\cdot 2 + 10$

$11 = 11\cdot 1 + 10$

Thus, $1 = 11 - 10 = 11 - (32 - 11\cdot 2) = 3\cdot 11 - 32$ .

Then $19 = 57\cdot 11 - 19\cdot 32$ .

The initial solution for $n$ is $57$ , and since $gcd(11,32) = 1$ , $n + 32p$ will also be the solutions where $p$ is some integer.

Since $57 \equiv 25 \pmod{32}$ . The least possible $n$ is $\boxed{25}$ .

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Checking for solutions: $75\cdot 25 \equiv 1875 \equiv 1856 + 19 \equiv 58\cdot 32 + 19 \equiv 19 \equiv 83 \pmod{32}$ .

From the statement, $83\equiv75n$ (mod 32), we can derive:

$\begin{cases} 83\mod 32 = r \\ 75\mod 32 = r \end{cases} \Rightarrow \begin{cases} 83 = 32k + r \\ 75 = 32l + r \end{cases} \quad \text{with l, r some positive integers} \quad \Rightarrow \quad 83-32k=75n-32l$

And thus we obtain that: $75n + 32(k-l)=83$ . We see that $gcd(75,32)=1$ . Because $1\mid 83$ we can try solving the simpler problem:

$75x^*+32y^*=1 \quad x^*, y^* \in \mathbb{Z}$

Using the Euclidean algorithm we can write:

$\begin{array}{ c|c|l} 75 = 2*32+11 & 11 = 75-2*32 & 1 = 11-10 = 11 - (32 - 2*11) = 3*11-32 \\ 32 = 2*11+10 & 10=32-2*11 & = 3*(75-2*32)-32 \\ 11 = 10*1 +1 & 1=11-10 & = 3*75+(-7)*32 \\ \end{array}$

So a potential solution is: $n=3*83=249, k-l=-7*83=-581$ . Using the general solutions formula we get:

$(249+32m, -581-75m)$

We can see: $\text{The smallest possible n is for } m=-7 \Rightarrow n=25$