\[79^{79} \equiv N \pmod{100}\]\(N\) is a two digit number.

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\[79^{79} \equiv N \pmod{100}\]\(N\) is a two digit number.

Mathematics
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Fermat's Little Theorem?
\[79 \equiv 79 \equiv 0 \pmod{79} \]Which is kinda obvious.
And so we meet again.

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Euler's Theorem says that \(\phi(79) = 79\) T_T
Yeah ;-)
Yeah? I was talking to the modular maths problem XD LOL Jk
So anyway, 79^5 = -1(mod 100)
How do you know again...
79^2 = 6241 = 41(mod 100) 79^3 = (41)(79)(mod 100) = 39(mod 100) 79^4 = (39)(79)(mod 100) = 81(mod 100) 79^5 = (81)(79)(mod 100) = 99(mod 100) = -1(mod 100)
So \(79^{10} \equiv 1 \pmod{100}\)?
Yeah, that's true....
Meaning 79^70 = 1(mod 100), too :D
Okay, oh.
So does that mean \(79^{80} \equiv 1 \pmod {100}\)
Ah, yeah.
Yes... But it's 79^79 you want, right? 79^75 = -1(mod 100)
\[79^{70} \times 79^{5} \times 79^{4} \pmod{100}\]AH!
mhmm... 79^70 is 1 79^5 is -1 79^4, refer to the process we did earlier...
\[1 \times -1 \times 81 \pmod{100}\]
Yeah, that seems right :D
\[19\pmod{100}\]
Bingo... :|
:-D

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