anonymous
  • anonymous
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?
Mathematics
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SOLVED
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chestercat
  • chestercat
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ParthKohli
  • ParthKohli
\[21^{15} \cdot 21^4 \equiv 1 \cdot 81 \equiv 81 \pmod{100} \]
anonymous
  • anonymous
Thanks, but I'm looking for something more general. Like: \[p^ {n+1} = p \mod 100\] Where p^n = 1 mod 100
anonymous
  • anonymous
\[21==21\]

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More answers

ParthKohli
  • ParthKohli
You have no way but to use the modulos.
anonymous
  • anonymous
\[21^2==41\]
anonymous
  • anonymous
\[21^3==61\]
ParthKohli
  • ParthKohli
\[21^4 \equiv 81\]
ParthKohli
  • ParthKohli
Yes, that is what I used to determine all the mods: pattern recognition. :-)
anonymous
  • anonymous
no until you reach 1 later on you can get the powr of 1
ParthKohli
  • ParthKohli
And write it in some crappy notation so that people think you are smart.
ParthKohli
  • ParthKohli
Repeats every four terms.
anonymous
  • anonymous
\[21^5=1\]
anonymous
  • anonymous
Well what if I have something a bigger repetition pattern? I don't want to brute force the problem.
ParthKohli
  • ParthKohli
You just have to find a power which is \(1\) or \(-1\) that thing. (thanks @terenzreignz!)
ParthKohli
  • ParthKohli
\(1\) or \(-1\) modulo*
ParthKohli
  • ParthKohli
You always have Fermat's Little Theorem!
terenzreignz
  • terenzreignz
Cute :) Then again, you could also express it as (20 + 1)^20 (mod 100) If that's any easier :D
ParthKohli
  • ParthKohli
\[a^{b - 1} \equiv 1 \pmod{b}\]

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