A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 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
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?

This Question is Closed

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

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

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

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

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

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

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

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

terenzreignz
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2\[a^{b  1} \equiv 1 \pmod{b}\]
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.