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Q. The last digit of number $7^{886}$ is

KCETKCET 2012Binomial Theorem

Solution:

Since, $ 7^{1}=7 $
$7^{2}=49,7^{3}=343,7^{4}=2401 $
$\therefore 7^{886}=\left(7^{4}\right)^{221} 7^{2}$
$\therefore $ The last digit number $7^{886}$ is $9 $