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 $17^2=289=290-1$
$\Rightarrow (17{)}^{256}={\left(17^2\right)}^{128}=(290-1{)}^{128}$
$={}^{128}{C}_0(290{)}^{128}-{}^{128}{C}_1(290{)}^{127}+{}^{128}{C}_2(290{)}^{126}-\dots -{}^{128}{C}_{125}(290{)}^3$
$+{}^{128}{C}_{126}(290{)}^2-{}^{128}{C}_{127}(290{)}^1+{}^{128}{C}_{128}(290{)}^0$
$=1000m+\frac{(128)(127)}{2}(290{)}^2-(128)(290)+1$
where, $m$ is a positive integer.
$=1000m+(128)(290)[(127)(145)-1]+1$
$=1000m+(128)(290)(18414)+1=1000m+683527680+1$
$=1000[m+683527]+680+1=1000[m+683527]+681$