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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
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

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

BenjaminEEBest 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

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

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

cahitBest 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

ParthKohliBest 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

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

BenjaminEEBest 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

ParthKohliBest 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

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

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

terenzreignzBest 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

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