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Take last digit of 97 i.e. 7

Now $7\times 7=9, 9\times 7=3,3\times 7=1,1\times 7=7$ (taking last digit only)

So, 7 is a cycle of 4 and after multiplication of four 7 we get 7 again. i.e., $7^{4n+1}$ will have unit digit as 7 for $n \geq 0$.

$2013=503\times 4+1$

Now, last digit of $97^{2013}$

= Last digit of  $7^{503\times 4+1}  =7.$

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