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BenjaminEE

  • 3 years ago

Modular arithmetic. 21^20 = 1 (mod 100) From here it's simple to get that 21^(20+1) = 21 (mod 100) But what about 21^(20-1) = ? (mod 100) Is there a simple method for this?

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  1. ParthKohli
    • 3 years ago
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    \[21^{15} \cdot 21^4 \equiv 1 \cdot 81 \equiv 81 \pmod{100} \]

  2. BenjaminEE
    • 3 years ago
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    Thanks, but I'm looking for something more general. Like: \[p^ {n+1} = p \mod 100\] Where p^n = 1 mod 100

  3. cahit
    • 3 years ago
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    \[21==21\]

  4. ParthKohli
    • 3 years ago
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    You have no way but to use the modulos.

  5. cahit
    • 3 years ago
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    \[21^2==41\]

  6. cahit
    • 3 years ago
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    \[21^3==61\]

  7. ParthKohli
    • 3 years ago
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    \[21^4 \equiv 81\]

  8. ParthKohli
    • 3 years ago
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    Yes, that is what I used to determine all the mods: pattern recognition. :-)

  9. cahit
    • 3 years ago
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    no until you reach 1 later on you can get the powr of 1

  10. ParthKohli
    • 3 years ago
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    And write it in some crappy notation so that people think you are smart.

  11. ParthKohli
    • 3 years ago
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    Repeats every four terms.

  12. cahit
    • 3 years ago
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    \[21^5=1\]

  13. BenjaminEE
    • 3 years ago
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    Well what if I have something a bigger repetition pattern? I don't want to brute force the problem.

  14. ParthKohli
    • 3 years ago
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    You just have to find a power which is \(1\) or \(-1\) that thing. (thanks @terenzreignz!)

  15. ParthKohli
    • 3 years ago
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    \(1\) or \(-1\) modulo*

  16. ParthKohli
    • 3 years ago
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    You always have Fermat's Little Theorem!

  17. terenzreignz
    • 3 years ago
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    Cute :) Then again, you could also express it as (20 + 1)^20 (mod 100) If that's any easier :D

  18. ParthKohli
    • 3 years ago
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    \[a^{b - 1} \equiv 1 \pmod{b}\]

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