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3)Find the last non-zero digit of $30^{2345}$

  • $3^1=3$
  • $3^2=9$
  • $3^3=27$
  • $3^4=81$
  • $3^5=243$ ... as last digit is following a cycle of $4$ so $2345/4$ gives remainder of 1, and $3^1=3$, so the last non-zero digit is 3.

This solution confuses me. Why can they use the powers of $3$ to determine the last digit of the $2345$th entry in the series of $30$?

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up vote 2 down vote accepted

The important point is that you are looking for the last non-zero digit - any additional multiples of 10 would merely tack a zero to the end.

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Oh my god this suddenly seems so obvious. It always is when someone's pointing at the answer... – Aerovistae Dec 20 '12 at 14:58

$30^n=3^n\cdot 10^n$, but since you are interested in the last non-zero digit, you need not consider $10^n$ (since $30^n$ is just $3^n$ with $n$ following $0$s).

Besides, $3^n$ may be recursively defined as $$ \begin{cases} 3^0 = 1 \\ 3^n = 3\cdot3^{n-1} \quad n\geq 1 \end{cases} $$ It is trivial that the last digit of $3^n$ is never $0$, and that it depends only on that of $3^{n-1}$; to be more precise, if the last digit of $3^{n-1}$ is $d$, then the last digit of $3^n$ is the last digit of $3d$. From the list of computations you provided, by a trivial induction it follows that the only possibilities for the last digit of $3^n$ are four, namely 1,3,9,7, and these occur recursively in this order.

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