#a way to calculate remainder of big numbers
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Do you know the Chinese reminder theorem?
Is it the system of equations stuff?
That's what i remember it as the last time I looked it up
Network Security: The Chinese Remainder Theorem (Solved Example 1)
Topics discussed:
- Chinese Remainder Theorem (CRT) statement and explanation of all the fields involved in the theorem.
- Explanation of working of CRT with different moduli that are relatively prime.
- Solved Example 1 on finding the common variable ‘X’ using CRT with 3 diff...
Did you understand it?
I think so but I couldn't remember it well since that was like a month ago
That time I was just looking up random stuff that I got interested in
I'll watch an explanation again, then see if I can solve it on my own, thanks
@echo stream question
Yes?
Am I suppose to apply the crt to solve it?
Apply to what?
For big numbers
Or do I watch it just to understand some stuff needed for solving it
For some questions you may have to apply a variant but usually you can do it with just crt.
I see
Better to understand the concept.
I am a bit rusty in crt as well, so I am also learning lol.
Oh thanks!
It doesn't actually have anything to do with the Chinese Remainder Theorem. You need to learn Euler's Theorem, or its more restricted form Fermat's Little Theorem.
The Chinese Remainder Theorem is about systems of congruences, but we don't have a system here.
Oh
I was referring to modular arithmetic.
Oh wait nvm I got it.
Ah a^p = a(mod p)
Sorry for the mishap.
What do you mean
Also, a good place to start would be reducing the base.
On the right hand side?
...there's only one side.
Huh?
What are the sides here?
When I tried solving this stuff I just compare the left side with something else then exponent it like 5 congruence to 1 (mod 4)
So 5^3 congruence to 1^3 (mod4)
Does it not work here?
What "left side"?
What n? Where did n come from?
...too big to... have n or...???
What I tried was like adding another side for 2000^n congruence to (-2)^n (mod 7) but it's too big
...why would it be congruent to that?
Also that still doesn't explain to me where n came from.
Isn't 2000 congruent to 5?
Nevermind
Okay, look.
We have one thing.
2000^2000 mod 7.
It's not an equation.
it's not a congruence.
There's no "left side" or "right side".
It's just an expression.
We're evaluating the expression.
Depends on what it is.
I did something like this
I think all I have to do is just use an exponent that makes it so the other side is 1,-1 but I'm not sure if it'll always work for big numbers
Thank you btw @cold nacelle @echo stream
@civic trellis has given 1 rep to @echo stream @cold nacelle
...I mean. Fermat's Little Theorem.
What did I do 
I'll read it, thx!