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Q. Last three digits of $17^{256}$ is

NTA AbhyasNTA Abhyas 2022

Solution:

Last three digits of $17^{256}$
$=\left(289\right)^{128}=\left(- 1 + 290\right)^{128}$
$=^{128}C_{0}\left(- 1\right)^{128}-^{128}C_{1}\cdot 290+^{128}C_{2}\cdot \left(290\right)^{2}+.......$
$=1-128\times 290+64\times 127\times 29^{2}\times 100+.......$
$=1-......120+.....800+.......$
$=681$