Find a $x$ such that $2^{2015}x\equiv 1 \pmod{13}$

Question

Since 13 is prime number using little Fermat's theorem $2^{12}\equiv 1 \pmod {13}$ then $2^{2015}\equiv 2^{12\cdot167+11}\equiv 2^{11} \pmod{13}$ then $2^{2015} x \equiv 2^{11} x \equiv 1 \pmod{13}$ so then $x\equiv 2$, is this ok?

Answer 1

By Fermat's theorem, $$2^{12}\equiv 1\pmod{13}$$ Observe that, ${12}\mid {2016}$ Therefore, $$ 2^{2016}\equiv 1\pmod{13}$$ Which is same as $$2^{2015}\times 2\equiv 1\pmod{13}$$ So $x= 2$ is a solution of the congruence.