What are the last two digits of $3844^{47}$ ?

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Solution

The correct option is D
64

$(3844)^{47} = (. . .44)^{47} = (11 \times 4)^{47} = (11)^{47} \times (2)^{94} = . .71 \times (2)^{90} \times 2^{4} = . .71 \times . .24 \times 16$
( $∵ (2^{10})^{o d d n u m b e r}$ always ends in 24)
The last two digits of the above product = 64.

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