Solving $315 x \equiv 5 \pmod {11}$ I have to solve this: $$315 x \equiv 5 \pmod {11}$$
Isn't it like that?
$$315 \equiv (22+8) \cdot 10+15 \equiv 8 \cdot 3+4 \equiv 5+8+4 \equiv 6$$
Or have I done something wrong?

If we use Fermat's little theorem and the fact that $315 \equiv 7\text{ (mod 11)}$, then $7^{11 - 1}\equiv 1 \text{ (mod } 11) \implies 7^{11 - 1} \cdot 5 \equiv 5 \text{ (mod } 11) \implies 7 \cdot 7^{11-2} \cdot 5\equiv 5\text{ (mod 11)} \implies 7 \cdot 8 \cdot 5 \equiv 5 \text{ (mod 11)} \implies x \equiv 40 \text{ (mod 11)} \implies x \equiv 7 \text{ (mod 11)}.$
