Question

The last three digits of $17^{256}$ are

• A. $680$
• B. $681$
• C. $689$
• D. None of these

Answer 1

Answer: (B)

To find the last three digits of $17^{256}$, consider $17^2=289=290-1$
$\Rightarrow(17)^{256}=\left(17^2\right)^{128}=(290-1)^{128}$
$={ }^{128} C_0(290)^{128}-{ }^{128} C_1(290)^{127}+{ }^{128} C_2(290)^{126}-\ldots-{ }^{128} C_{125}(290)^3$
$+{ }^{128} C_{126}(290)^2-{ }^{128} C_{127}(290)^1+{ }^{128} C_{128}(290)^0$
$=1000 m+\frac{(128)(127)}{2}(290)^2-(128)(290)+1$
where, $m$ is a positive integer.
$=1000 m+(128)(290)[(127)(145)-1]+1 $
$=1000 m+(128)(290)(18414)+1=1000 m+683527680+1 $
$=1000[m+683527]+680+1=1000[m+683527]+681$