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?

Question

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?

parthkohli · Answer

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

anonymous · Answer

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

anonymous · Answer

$$21==21$$

parthkohli · Answer

You have no way but to use the modulos.

anonymous · Answer

$$21^2==41$$

anonymous · Answer

$$21^3==61$$

parthkohli · Answer

$$21^4 \equiv 81$$

parthkohli · Answer

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

anonymous · Answer

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

parthkohli · Answer

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

parthkohli · Answer

Repeats every four terms.

anonymous · Answer

$$21^5=1$$

anonymous · Answer

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

parthkohli · Answer

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

parthkohli · Answer

$1$ or $-1$ modulo*

parthkohli · Answer

You always have Fermat's Little Theorem!

terenzreignz · Answer

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

parthkohli · Answer

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