Question

Find the last two digits of ${56}^{283}$.

Answer 1

${56}^{283}=(2^3\times 7{)}^{283}=2^{849}\times 7^{283}=(2^{10}{)}^{84}\times 2^9\times (7^4{)}^{70}\times 7^3=76\times 12\times (01{)}^{70}\times 43=16$