Question

Find the last two digits of ${87}^{474}$.

Answer 1

${87}^{474}={87}^{472}\times {87}^2=({87}^4{)}^{118}\times {87}^2=(69\times 69{)}^{118}\times 69$ [$\therefore$ The last two digits of ${87}^2$ are 69]
= ${61}^{118}\times 69=81\times 69=89$