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Q.
The last two digits of $17^{256}$ is
Binomial Theorem
Solution:
$17^{256}=\left(17^2\right)^{128}=(290-1)^{128}$
$=1000 m +{ }^{128} C _2(290)^2-{ }^{128} C _1(290)+1$
$=1000( m +683527)+681$
$\therefore$ last two digits $=81$