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BenjaminEE
Group Title
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^(201) = ? (mod 100)
Is there a simple method for this?
 one year ago
 one year ago
BenjaminEE Group Title
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^(201) = ? (mod 100) Is there a simple method for this?
 one year ago
 one year ago

This Question is Closed

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
\[21^{15} \cdot 21^4 \equiv 1 \cdot 81 \equiv 81 \pmod{100} \]
 one year ago

BenjaminEE Group TitleBest ResponseYou've already chosen the best response.0
Thanks, but I'm looking for something more general. Like: \[p^ {n+1} = p \mod 100\] Where p^n = 1 mod 100
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
You have no way but to use the modulos.
 one year ago

cahit Group TitleBest ResponseYou've already chosen the best response.0
\[21^2==41\]
 one year ago

cahit Group TitleBest ResponseYou've already chosen the best response.0
\[21^3==61\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
\[21^4 \equiv 81\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Yes, that is what I used to determine all the mods: pattern recognition. :)
 one year ago

cahit Group TitleBest ResponseYou've already chosen the best response.0
no until you reach 1 later on you can get the powr of 1
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
And write it in some crappy notation so that people think you are smart.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Repeats every four terms.
 one year ago

BenjaminEE Group TitleBest ResponseYou've already chosen the best response.0
Well what if I have something a bigger repetition pattern? I don't want to brute force the problem.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
You just have to find a power which is \(1\) or \(1\) that thing. (thanks @terenzreignz!)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
\(1\) or \(1\) modulo*
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
You always have Fermat's Little Theorem!
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Cute :) Then again, you could also express it as (20 + 1)^20 (mod 100) If that's any easier :D
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
\[a^{b  1} \equiv 1 \pmod{b}\]
 one year ago
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