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Q. Last two digits of $21^{100}$ are

Binomial Theorem

Solution:

$(21)^{100} =(1+20)^{100} $
$=1+{ }^{100} C _{1} \cdot 20+{ }^{100} C _{2} \cdot 20^{2}+\cdots+{ }^{100} C _{100} \cdot 20^{100} $
Hence, last two digit are $01$ .