Question

Find the last $3$ digit of no $7^{51}$.

Answer 1

$\Rightarrow 7^{51}=7\left(7^{50}\right)$

$\Rightarrow 7\left(\right(7^2{)}^{25})$

$7\left({49}^{25}\right)=7(50-1{)}^{25}$

$\Rightarrow 7\left({}^{25}{C}_0\right(-1)+{}^{25}{C}_1(-1{)}^2\left(50\right)+.........)$

$\Rightarrow (-1+25(+1\left)\right(50):-)$

$\Rightarrow (-1+(25\left)\right(50\left)\right)$

$\Rightarrow 7\left(1249\right)\Rightarrow 8743$

So last $3$ digit are :- $743$