What is the remainder when (37)2012 is divided by 25?

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Question

What is the remainder when

(37)^{2012}

is divided by 25?

Answer: Option A
:
A
Last two digits of

(37)^{2012} = (69)^{1006} = (61)^{503} = \frac{81}{25} ∣ r = 6

.
Note - The ten's digit of

61^{n}

has a cyclicity of 5 (61-21-81-41-01-61) and 503=5k+3.

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