Thank you for reporting, we will resolve it shortly
Q.
The last digit of the number $(32)^{32}$ is
Binomial Theorem
Solution:
$(32)^{32} =(2+3 \times 10)^{32} $
$=2^{32}+10 k, $ where $ k \in N$
Therefore, last digits in $(32)^{32}=$ last digit in $(2)^{32}$
But $2^{1}=2,2^{2}=4,2^{3}=8,2^{4}=16,2^{5}=32$
$\therefore 2^{32}-\left(2^{5}\right)^{6} \cdot 2^{2} =(32)^{6} \cdot 4=(2+30)^{6} \cdot 4$
$=\left(2^{6}+10 r\right) 4, r \in N$
Last digit in $2^{32}=$ last digit in $(2)^{6} \cdot 4=$ last digit in $4 \times 4=6$
$\therefore $ Last digit in $(32)^{32}=6$.