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ParthKohli

  • 2 years ago

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

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  1. ParthKohli
    • 2 years ago
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    Fermat's Little Theorem?

  2. ParthKohli
    • 2 years ago
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    \[79 \equiv 79 \equiv 0 \pmod{79} \]Which is kinda obvious.

  3. terenzreignz
    • 2 years ago
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    And so we meet again.

  4. ParthKohli
    • 2 years ago
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    Euler's Theorem says that \(\phi(79) = 79\) T_T

  5. ParthKohli
    • 2 years ago
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    Yeah ;-)

  6. terenzreignz
    • 2 years ago
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    Yeah? I was talking to the modular maths problem XD LOL Jk

  7. terenzreignz
    • 2 years ago
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    So anyway, 79^5 = -1(mod 100)

  8. ParthKohli
    • 2 years ago
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    How do you know again...

  9. terenzreignz
    • 2 years ago
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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)

  10. ParthKohli
    • 2 years ago
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    So \(79^{10} \equiv 1 \pmod{100}\)?

  11. terenzreignz
    • 2 years ago
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    Yeah, that's true....

  12. terenzreignz
    • 2 years ago
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    Meaning 79^70 = 1(mod 100), too :D

  13. ParthKohli
    • 2 years ago
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    Okay, oh.

  14. ParthKohli
    • 2 years ago
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    So does that mean \(79^{80} \equiv 1 \pmod {100}\)

  15. ParthKohli
    • 2 years ago
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    Ah, yeah.

  16. terenzreignz
    • 2 years ago
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    Yes... But it's 79^79 you want, right? 79^75 = -1(mod 100)

  17. ParthKohli
    • 2 years ago
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    \[79^{70} \times 79^{5} \times 79^{4} \pmod{100}\]AH!

  18. terenzreignz
    • 2 years ago
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    mhmm... 79^70 is 1 79^5 is -1 79^4, refer to the process we did earlier...

  19. ParthKohli
    • 2 years ago
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    \[1 \times -1 \times 81 \pmod{100}\]

  20. terenzreignz
    • 2 years ago
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    Yeah, that seems right :D

  21. ParthKohli
    • 2 years ago
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    \[19\pmod{100}\]

  22. terenzreignz
    • 2 years ago
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    Bingo... :|

  23. ParthKohli
    • 2 years ago
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    :-D

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