Question

Find the last two digits of the number ${\left(17\right)}^{10}.$

Answer 1

${\left(17\right)}^{10}={\left(289\right)}^5={(290-1)}^5$
${=}^5{\text{C}}_0{\left(290\right)}^5{-}^5{\text{C}}_1{\left(290\right)}^4+...{+}^5{\text{C}}_4{\left(290\right)}^1{-}^5{\text{C}}_5{\left(290\right)}^0$
${=}^5{\text{C}}_0{\left(290\right)}^5{-}^5{\text{C}}_1{\left(290\right)}^4+...{+}^5{\text{C}}_3{\left(290\right)}^2+5\times 290-1$
=A multiple of $1000+1449$
Hence, last two digits are $49$