How can I find the second last digit of the number $7^{41^{17}+1}$? I'm not quite sure how to start. A hint would suffice. Thanks!
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work mod $100$. Euler's theorem can be helpful. – Anurag A Jun 27 '17 at 23:34
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$7^4 = 2401$ is a helpful bit of knowledge to have for this problem. – Doug M Jun 27 '17 at 23:56
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See: https://math.stackexchange.com/questions/362012/find-the-last-two-digits-of-781 – lab bhattacharjee Jun 28 '17 at 06:38
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You need to calculate $7^{41^{17}+1} \pmod {100}$ then read off the leading digit from the result. You do that the same way you did in your other question. Euler's theorem again.
Ross Millikan
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Let's see ...
$7^1=7, 7^2=49, 7^3=343, 7^4=2401, ...$
Hmmm, of $7^4$ ends with $01$, what does that say about $7^8, 7^{12}, 7^{any.multiple.of.4}$?
$41^{17}-1$ is easily seen to be a multiple of $4$. Then $7$ to that power ends with $01$ and multiying by $7^2=49$ implies $7^{41^{17}+1}$ ends with ... .
Got it?
Oscar Lanzi
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