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



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ParthKohli
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Fermat's Little Theorem?

ParthKohli
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\[79 \equiv 79 \equiv 0 \pmod{79} \]Which is kinda obvious.

terenzreignz
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And so we meet again.

ParthKohli
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Euler's Theorem says that \(\phi(79) = 79\) T_T

ParthKohli
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Yeah ;)

terenzreignz
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Yeah? I was talking to the modular maths problem XD
LOL
Jk

terenzreignz
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So anyway, 79^5 = 1(mod 100)

ParthKohli
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How do you know again...

terenzreignz
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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
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So \(79^{10} \equiv 1 \pmod{100}\)?

terenzreignz
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Yeah, that's true....

terenzreignz
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Meaning 79^70 = 1(mod 100), too :D

ParthKohli
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Okay, oh.

ParthKohli
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So does that mean \(79^{80} \equiv 1 \pmod {100}\)

ParthKohli
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Ah, yeah.

terenzreignz
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Yes... But it's 79^79 you want, right?
79^75 = 1(mod 100)

ParthKohli
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\[79^{70} \times 79^{5} \times 79^{4} \pmod{100}\]AH!

terenzreignz
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mhmm...
79^70 is 1
79^5 is 1
79^4, refer to the process we did earlier...

ParthKohli
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\[1 \times 1 \times 81 \pmod{100}\]

terenzreignz
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Yeah, that seems right :D

ParthKohli
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\[19\pmod{100}\]

terenzreignz
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Bingo... :

ParthKohli
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:D