A "fast" approach to solve $2^{133} \equiv x \mod 133 $ I have to solve this equation $2^{133} \equiv x \mod  133 $.Using Euler's theorem I reduced it to $2^{25} \equiv x \mod 133$ but I couldn't think off any fast way to proceed after this.
Any ideas?

Use $133 = 7 \times 19$. Use $2^p = 2 \mod p$ for $p \in \mathbb{P}$.
Now $2^{133} = (2^7)^{19} \mod 7 = 2^{19} \mod 7 = (2^7)^2 \times 2^5 = 2^2 \times 2^5 = 2^7 = 2 \mod 7$. Similarly $2^{133} = (2^{19})^7 = 2^7 = 14 \mod 19$. Thus
Let $x = 2^{133} \mod 133$, then from $x = 14 \mod 19$, it follows that  $x = 14 + 19 \times k$. From $x = 2 \mod 7$, it follows $(14 + 19 \times k) = 5 \times k = 2 \mod 7$, and $k = 6 \mod 7$. From this I get $2^{133} = 128 \mod 133$.
