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I'm stuck on this problem:

(1) In the sequence $7,7^2,7^3,7^4,\ldots,7^{2014}$ how many terms have $3$ as the units digit?

After some random stuff, I have found that the unit digits of $7$ go in the order $7,9,3,1$ And then back to $7$. But I don't know how to utilize this in the problem!

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1 Answer 1

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You are doing it correctly. Simply divide $2014$ by $4$, where you get $503$, plus a $7^{2013}, 7^{2014}$ at the end. Since the $3$ comes $3rd$ in the sequence, it will not appear in the last two, so the answer is just $503$.

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    $\begingroup$ Brilliant! Thanks! Just what I needed. $\endgroup$
    – user332252
    Commented Oct 2, 2016 at 17:53

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