What is the remainder when 3^99 is divided by 8?

3^99 can be re-written as (4-1)^99
Let’s call 4 as x,and -1as y.

When we do the binomial expansion of (x+y)^99, and temporarily omit the coefficient, we obtain,
X^99+x^98*y+x^97*y^2+x^96*y^3x*y^98+y^99.

We know that 4 raised to any natural number grater than one,will divide 8 completely (4^2=16=8*2, and therefore every subsequent power of 4 can be rewritten in the form of 8k, where k is a natural number.)

The last two terms of this binomial expansion are,
(99 c 98) *x*y^98 + y^99(x C y represents x “choose” y)
99 C 98 is nothing but 99 ( by using the n C r formula )
So the last two terms will be,
99*4*1+(-1)
=396-1=395
395 mod 8= (392 mod 8) + (3 mod 8) = 0+3=3
So our required answer has been obtained: 3

Hope that helps

B. 16is the absolute value