Question

find the remainder when 3^75 is devided by 5.

Answer 1

hey there,

Here's how I figured it out:
3^1 = 3, divided by 5 creates a remainder of 3.
3^2 = 9, divided by 5 creates a remainder of 4.
3^3 = 27, div. by 5 creates a remainder of 2.
3^4 = 81, div. by 5 creates a remainder of 1.
----(REMAINDERS of 5 can only be 1-4, so now this process repeats itself.)
so 3^5 = 243, which is a remainder of 3.
so
3^(1+ 4n) :: remainder = 3
3^(2+ 4n) :: remainder = 4.
3^(3+ 4n) :: remainder = 2.
3^(4+ 4n) :: remainder = 1.

So, to get 3^98, we need to find how to get 98 with our equations.
(4 *24 = 96)
It ends up being 2+96 = 98, so we have the equation
3^(2 + 4n) :: remainder = 4.

Our remainder will be 4.

Hope this helps!