Simplifying the modulus of a linear congruences

Question

I have seen the rule that if $\gcd(k,m) =1 $ and we have $ka \equiv kb \pmod{m}$ then $a \equiv b \pmod{m}$. But I have just seen that $3(7k) \equiv 3 \pmod{15} \implies 2k \equiv 1 \pmod{5}$ What rule is being used here? Why has the modulus also been divided by 3? Is it to do with the fact 15 isn't prime?

Answer 1

In the following implication, $$3(7k) \equiv 3 \space (mod \space 15) \implies 2k \equiv 1 \space (mod \space 5)$$

Note that if $$3(7k) \equiv 3 \space (mod \space 15)$$ then for some integer $m$, we have $$3(7k)-3=15m$$ and we can divide by three to get $$7k-1=5m$$ which implies $$ 2k \equiv 1 \space (mod \space 5)$$

Answer 2

Note that $\gcd(3, 15) = 3 \neq 1$ in this case, so you cannot use that rule. In general, if $ka \equiv kb \pmod m$ then $a \equiv b \pmod {\frac{m}{\gcd(k,m)}}$, hence why you must divide by $3$ here.