Question

Day 7!

This is gonna be a really hard one, and I will be very surprised if someone figures this one out. It is gonna be a mathematics problem, and i want you to figure out the logic.

2 can be written as 3n+ 2 or 3n - 1, where n is any number, that doesn't really matter.
4 can be written as 3n + 1, and so on.
Here, the second term is the remainder, when 2 and 4 are divided by three, and n is the quotient.

So my question for day 7 is this: What is the remainder when 2^100 is divided by three?
(Difficulty: Very hard)

I am awarding a lot of points, doesn't mean you will spam random letters or 'Hellos' Or 'Good afternoon' or anything else other than the answer and explanation. I report all such pointless answers.
Good luck!

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Answer 1

Answer:

As 2^100 is a multiple of 4 the answer is 1

Answer 2

Fermat's little theorem states that if p is a prime number, then for any integer a, the number a^p − a is an integer multiple of p. In the notation of modular arithmetic, this is expressed as a (formula given in the attachment)