Modular exponentiation with exponent that has repeating pattern

Question

If i have to count a large exponent modulo N, that has some repeating pattern. Is there some way how to simplify the exponent before actually counting the remainder?

$131^{123456789876543212345678987654321 } \;(mod \;12)$

Answer 1

$131=120+11 \equiv 11 \equiv-1\;(\bmod\; 12)$. As the exponent is odd then $131^{123456789876543212345678987654321} \equiv-1\;(\bmod\; 12)$, this is a way of simplifying the basis before counting the remainder.