What are the last two digits for $6^{1000}$? And how would I figure out other numbers (such as $73, 283$, etc.) to the $1000$th power?

Question

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.

Answer 1

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 $73$ ...

Answer 2

$6\equiv 6 \pmod{100}\\ 6^2\equiv 36 \pmod{100}\\ 6^3\equiv 16 \pmod{100}\\ 6^4\equiv 96 \pmod{100}\\ 6^5\equiv 76 \pmod{100}\\ 6^6\equiv 56 \pmod{100}\\ 6^7\equiv 36 \pmod{100}$

and from there on out the cycle repeats.

When $n>1$

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

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