What is the remainder when $(37)^{2012}$ is divided by 25?

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Solution

The correct option is A
31

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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