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Q. The remainder when $27^{10}+7^{51}$ is divided by $10$

Binomial Theorem

Solution:

$(27)^{10}+7^{51} =(30-3)^{10}+(10-3)^{51}$
$=3^{10}-3^{51}+10 \lambda $
$=(10-1)^{5}-3(10-1)^{25}+10 \lambda $
$=-1+3+10 \lambda_{1} $
$=2+10 \lambda_{1} $
$\therefore $ Remainder is 2