I was working on a problem that resulted in the calculation:
$20 \equiv 10x \pmod{11}$.
I got the answer $x \equiv 2 \pmod{11}$ with the thought process: Since $10$ is a factor of $20$ I can rewrite $20 \pmod{11}$ as $10x \pmod{11}$ with $x=2$. But isn't this basically the same as using division, which we aren't supposed to do in modular spaces?
If I solve by multiplying $20$ by the multiplicative inverse of $10 \pmod{11}$ which is $10$, I get $200 \pmod{11}$ which simplifies to $2 \pmod{11}$ anyway. Did these numbers just happen to be the same or is my original method valid?
Answer
Your original method works only because $10$ has an inverse modulo $11$. You wrote $20 \equiv 10x \pmod{11}$ as
$$ 10(2) \equiv 10x \pmod{11}. $$
Now, using the fact that $10^{-1}$ exists, we can multiply by it to conclude $2 \equiv x \pmod{11}$. If $10$ wasn't invertible, we wouldn't be able to conclude this. For instance, $2(2) \equiv 2(3) \pmod{2}$, but $2\not\equiv 3 \pmod{2}$.
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