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Q. The last digit in $7^{300}$ is

Binomial Theorem

Solution:

We have, $7^{1}=7,7^{2}=7 \times 7=49$
$7^{3}=7 \times 7 \times 7=343 ; $
$7^{4}=7 \times 7 \times 7 \times 7=2401$
$7^{5}=7 \times 7 \times 7 \times 7 \times 7=16807$
Last digit of $7^{1}=7,7^{2}=9,7^{3}=3,7^{4}=1$ and $7^{5}=7$ thus cycle of last digit repeats at $7^{5} .$
$\therefore $ Last digit of $7^{300}=1$