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I think I have to get the pattern of repeating numbers and divide the exponent by the number of repeating numbers, but what if $1000$ is divisible by both $4$ and $5$, since the number of repeating numbers for the second last digit is $5$? Thank you so much in advance.

Abby
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  • Essentially, for looking at the $n$ last digits of some large power, observe the cycle the powers go through mod $10^n$. – Isky Mathews May 15 '18 at 21:34

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Yes, just try to find that repeating pattern for the last two digits:

$6, 36, 16, 96, 76, 56, 36, ...$

OK, we found our pattern! So it's cyclic with a period of $5$, and since $1000$ is divisible by $5$, we can just look at the $5$-th entry, so it's $76$

For other numbers, find the period of their cycles. And you know the period is $50$ at the most, since there are only $100$ different two-digit numbers, and $50$ of those are even, and $50$ are odd. Still, I'm glad you didn't ask me about $73$ ...

Bram28
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  • Note that powers of $2$ are divisible by $4$ except for the first, which halves the number of possibilities, and then $20, 40, 60, 80, 00$ will never occur so the possibilities reduce to $20$. With odd multiples of $5$ the possible repeating values are $25, 75$ only. For odd numbers coprime to $100$ there are $40$ possibilities. The periods are factors of $20$, $2$ and $40$ respectively. These observations generalise. – Mark Bennet May 15 '18 at 21:32
  • @MarkBennet Yeah, just found for $73$ the period is $20$ ... is there an easy way to see why these periods are like that? If so, I think that would make for a good Answer to the OP! – Bram28 May 15 '18 at 21:36
  • There is a result known as the Fermat-Euler Theorem which applies when numbers are co-prime. If they are not the results are not so well known. – Mark Bennet May 15 '18 at 21:52
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$6\equiv 6 \pmod{100}\\ 6^2\equiv 36 \pmod{100}\\ 6^3\equiv 16 \pmod{100}\\ 6^4\equiv 96 \pmod{100}\\ 6^5\equiv 76 \pmod{100}\\ 6^6\equiv 56 \pmod{100}\\ 6^7\equiv 36 \pmod{100}$

and from there on out the cycle repeats.

When $n>1$

$6^{(n+5)}\equiv 6^n \pmod{100}$

$6^{1000}\equiv 6^{995} \equiv 6^{990} \equiv 6^5 \pmod{100}$

Doug M
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  • $6^4 \equiv 16*6 \equiv 96 \not \equiv 76$. And $6^6 \equiv 36$ and $6^4\equiv 36$ would imply $6^{n+4}\equiv 6^n$. (This is two wrongs making a right). And $6^5 \equiv 76 \not \equiv 56$. (So this is three wrongs making a wrong.) – fleablood May 15 '18 at 21:24