ParthKohli
  • ParthKohli
\[79^{79} \equiv N \pmod{100}\]\(N\) is a two digit number.
Mathematics
jamiebookeater
  • jamiebookeater
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ParthKohli
  • ParthKohli
Fermat's Little Theorem?
ParthKohli
  • ParthKohli
\[79 \equiv 79 \equiv 0 \pmod{79} \]Which is kinda obvious.
terenzreignz
  • terenzreignz
And so we meet again.

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

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