Solving $472x ≡ 32 \;(\text{mod } 92)$: How to solve when $a$ is greater than $m$

Question

I'm not sure where to begin here. I think that I can change the $472x$ to be $12x$ because they should be equal in mod $92$. Then, I believe I can simplify the equation to be: $3x ≡ 8 \; (\text{mod } 23)$

Answer 1

You're on the right track. Starting from $$ 3x \equiv 8 \; (\text{mod } 23), $$ you should find the multiplicative inverse $3^{-1}$ of $3$ modulo $23$, i.e. the number satisfying $$ 3^{-1} \cdot 3 \equiv 1 \; (\text{mod } 23). $$ This can be done with the help of the Euclidean algorithm, or alternatively you could just notice that $$ 8\cdot 3 = 24 \equiv 1 \; (\text{mod } 23), $$ so $3^{-1} = 8$ in this case. Then you can multiply both sides of your equation by this multiplicative inverse to obtain $$ x \equiv 8\cdot 3 x \equiv 8 \cdot 8 = 64 \equiv 18 \; (\text{mod } 23). $$ Hence $x = 18+23k$, for an integer $k$, so you get four incongruent solutions (modulo $92$) to your original equation, namely by inserting $k=0,1,2,3$.