I know that
$$a \equiv b ~(\text{mod}~3)$$ and $$c \cdot a \equiv 1 ~(\text{mod}~3)$$
Can I substitute $a$ with $b$? I mean:
$$c\cdot b \equiv 1 ~(\text{mod}~3)$$
2026-04-08 09:33:56.1775640836
Modular arithmetic - is this a "legal" substitution?
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Yes!
We have $b = a + 3k$, so $$c\cdot b = c\cdot(a + 3k) = c\cdot a + c\cdot 3k \equiv c\cdot a (\mod 3)$$