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Q.
The last three digits of the number $(27)^{27}$ is
Binomial Theorem
Solution:
$ (27)^{27}=3^{81}=3 \cdot(9)^{40}$
$ =3(10-1)^{40}=3\left(10^{40}-{ }^{40} C _1 \cdot 10^{39}+\ldots . .+{ }^{40} C _{38} \cdot 10^2-{ }^{40} C _{39} \cdot 10+1\right) $
$=3(1000 \lambda-400+1)$
Last 3 digits of this number $=803$.