How can the equation like $$ x^{118}\equiv 113\;\; (mod\; 1001) $$ if I know the Fermat's little theorem, Chinese remainder theorem, Euler's theorem and basic operations on congruence?
My approach:
By Euler's theorem we know that if $x\perp 1001 $ then $x^{720}\equiv 1 \;\; (mod\; 1001)$ but cannot combine this with my equation.
Since $1001=7\cdot11\cdot13$ and $(113,1001)=1,$ we obtain $$x^6\equiv1(mod7),$$ $$x^{10}\equiv1(mod11)$$ and $$x^{12}\equiv1(mod13),$$ which since $[6,12,10]=60$, gives $$x^{60}\equiv1(mod1001)$$ and we need to solve $$x^{-2}\equiv113(mod1001).$$ Now, by the euclidean algorithm we can get that $$7\cdot1001=62\cdot113+1,$$ which gives $$62x^{-2}\equiv62\cdot113(mod1001)$$ or $$62x^{-2}\equiv-1(mod1001)$$ or $$x^2\equiv-62(mod1001)$$ or $$x^2\equiv18\cdot1001-62(mod1001)$$ or $$x^2\equiv134^2(mod1001).$$ Can you end it now?
You can get all eight solutions by the chinese remainder theorem.
We have the following system: $$x^2\equiv-62(mod7),$$ $$x^2\equiv-62(mod11)$$ and $$x^2=-62(mod13)$$ or $$x^2\equiv1(mod7),$$ $$x^2\equiv4(mod11)$$ and $$x^2=16(mod13)$$ or
$$x\equiv\pm1(mod7),$$ $$x\equiv\pm2(mod11)$$ and $$x\equiv\pm4(mod13).$$ We got eight systems and it's enough to solve four of them with $x\equiv1(mod7)$ for example.
We have first of these four systems: $$x\equiv1(mod7),$$ $$x\equiv2(mod11)$$ and $$x\equiv4(mod13).$$ Now, let $$143a\equiv1(mod7),$$ $$91b\equiv2(mod11)$$ and $$77c\equiv4(mod13).$$ It gives $$a\equiv5(mod7),$$ $$b\equiv8(mod11)$$ and $$c\equiv9(mod13).$$ Id est, $$x\equiv5\cdot143+8\cdot91+9\cdot77(mod1001)\equiv134(mod1001).$$ By the same way you can get other six solutions.