What is the remainder when (49)²⁰⁰⁸ is divisible  by 64 ?

By using the polynomial technique, it can be represented as (6x8 +1)²⁰⁰⁸  or (y+l)²⁰⁰⁸ where y is divisible by 8.

(y+1)ⁿ = yⁿ + nC₁y^n-1 + nC₂y^n-2 ………. nC_n-1y + 1

When n = 2,  the last but one term is 2y =2x6x8 ;on division by 64  the remainder can be obtained from 2x6x8/64 and the reduced remainder  from  2x6/8= 3/2  the reduced remainder is 1 and the actual remainder will be 32 timed the reduced remainder .Adding the last l  the resultant remainder will be 33.

When n = 3, the remainder is obtained from 3x6x8/64 =9/4, the reduced remainder = 1, and the actual remainder is 16 times  the reduced remainder i.e., 16 and adding the last 1 the resultant remainder is 17.

In the given problem the remainder can be obtained from 2008x6x8. It is divisible by 64 as 2008 is divisible by 8. Hence the reduced remainder will be zero. Adding the final 1, the final remainder will be 1. Hence if the power is divisible by 8, the remainder will always be equal to 1.