Modular Arithmetic - Changing a Congruence Modulo Relation

Question

Here the solution for $27^{17} \pmod{15}$ is shown.

Why is $27 \equiv 12 \pmod{15}$ rewritten as $27 \equiv -3 \pmod{15}$?

What is the logic/proof?

Answer 1

It can be proved by arguing that $15| (27-12)$ and $15|(27-(-3))$.

This is because in general, $$a\equiv b \pmod{n} \iff n|(a-b)$$

We can use this to show that $$a \equiv a-n \pmod{n}$$

This can be applied to your case : $27 \equiv 27-15=12 \pmod {15}$ and $12\equiv 12-15=-3\pmod{15}$