## 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^(20-1) = ? (mod 100) Is there a simple method for this? one year ago one year ago

1. ParthKohli

$21^{15} \cdot 21^4 \equiv 1 \cdot 81 \equiv 81 \pmod{100}$

2. BenjaminEE

Thanks, but I'm looking for something more general. Like: $p^ {n+1} = p \mod 100$ Where p^n = 1 mod 100

3. cahit

$21==21$

4. ParthKohli

You have no way but to use the modulos.

5. cahit

$21^2==41$

6. cahit

$21^3==61$

7. ParthKohli

$21^4 \equiv 81$

8. ParthKohli

Yes, that is what I used to determine all the mods: pattern recognition. :-)

9. cahit

no until you reach 1 later on you can get the powr of 1

10. ParthKohli

And write it in some crappy notation so that people think you are smart.

11. ParthKohli

Repeats every four terms.

12. cahit

$21^5=1$

13. BenjaminEE

Well what if I have something a bigger repetition pattern? I don't want to brute force the problem.

14. ParthKohli

You just have to find a power which is $$1$$ or $$-1$$ that thing. (thanks @terenzreignz!)

15. ParthKohli

$$1$$ or $$-1$$ modulo*

16. ParthKohli

You always have Fermat's Little Theorem!

17. terenzreignz

Cute :) Then again, you could also express it as (20 + 1)^20 (mod 100) If that's any easier :D

18. ParthKohli

$a^{b - 1} \equiv 1 \pmod{b}$