Question

Can you solve this ​

Answer 1

$\huge\bf { \red A \green N \pink S\blue W \orange E \purple R \green{=4}}$

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 it is helpful.

Answer 2

Step-by-step explanation:

mere answer ko BRAINList dijiye I am inbox you