can anyone show me the method to find the remainder of 2^2005 when divided by 3?

Question

please show me and explain to me in detail

Answer 1

consider

(3-1)^x =
3^x - A * 3^(x-1) * 1 + A * 3^(x-2) * 1^2 -..... + 1

or -1 in the end, if x is odd

all the terms, except that final +1 or -1 are divisible by 3, so remainder is 0

now if the last term is +1, remainder is 1. If it's -1, remainder is 2.

Note that for even x, its +1, for odd -1

so since 2005 is odd, the remainder is 2

Answer 2

2^1/3 = 2/3 = 0 remain 2
2^2/3 = 4/3 = 1 remain 1
2^3/3= 8/3 = 2 remain 2
2^4/3 = 16/3 = 5 remain 1
2^5/3 = 32/3 = 10 remain 2

This will continue, so the odd powers will have a remainder of 2, so 2005 should have a remainder of 2

Answer 3

2^2005
= (3-1)^2005
= (lots of terms with a factor of 3) + (-1)^2005
So the remainder when divided by 3 is the same as for (-1)^2005 = -1; add 3 to this to get a final answer of 2.

Formally, in modular arithmetic we write
2^2005 ≡ (-1)^2005 (mod 3)
≡ (-1) (mod 3)
≡ 2 (mod 3).