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BenjaminEE
 2 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^(201) = ? (mod 100)
Is there a simple method for this?
BenjaminEE
 2 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^(201) = ? (mod 100) Is there a simple method for this?

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

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2You have no way but to use the modulos.

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

cahit
 2 years ago
Best ResponseYou've already chosen the best response.0no until you reach 1 later on you can get the powr of 1

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2Repeats every four terms.

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

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2\(1\) or \(1\) modulo*

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2You always have Fermat's Little Theorem!

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2\[a^{b  1} \equiv 1 \pmod{b}\]
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