Question

The last two digits of $17^{256}$ is

• A. 18
• B. 81
• C. 71
• D. 17

Answer 1

Answer: (B)

$17^{256}={\left(17^2\right)}^{128}=(290-1{)}^{128}$
$=1000m+{}^{128}{C}_2(290{)}^2-{}^{128}{C}_1(290)+1$
$=1000(m+683527)+681$
$\therefore$ last two digits $=81$