Question

Find the last digit, last two digits and last three digits of the number $(27{)}^{27}$.

Answer 1

${\left(27\right)}^{27}={\left(3^3\right)}^{27}=3^{81}$

Now we know the cyclioity of $3$ is $4$. Hence, $81/4$ leaves a remainder of $1$.

Therefore, the last digit will be $3$.

Now,

$3^{81}=3\times 3^{80}=3\times {\left(3^2\right)}^{40}=3\times 9^{40}=3\times {(10-1)}^{40}=3\times ({10}^{40}{-}^{40}{C}_1\times {10}^{39}-....{-}^{40}{C}_{38}\times {10}^2{-}^{40}{C}_{39}\times 10+1)........\left(i\right)$

For the last two digits, divide the above expression by $100$. Each term of the above expression contains ${10}^2$ except $1$.

$\therefore 3^{81}=3\times (100\lambda +1)=300\lambda +3$

Therefore, the last two digits will be $03$

For the last three digits, divide equation (i) by $1000$. Each term of the above expression contains ${10}^3$ except ${-}^{40}{C}_{39}\times 10+1=-400+1=-399$

$\therefore 3^{81}=3\times (1000\lambda +(-399))=3000\lambda -1197$

Therefore, the last three digits will basically be the remainder of $-1197/1000=803$

Therefore, the last three digits are $803$.