Remainder????????????????????

this can be done like ,

8^333 / 7

i.e

(7 + 1)^333 / 7

this is

(7 + 1)(7 + 1)(7 + 1)..............(7 + 1) / 7

now when u can treat 7 as x...

(x + 1)(x + 1)....................(x + 1) / x

now when u open it all the terms will be a multiple of x i.e 7 except the last term ie 1.

so the multiples of 7 will not leave any remainder and 1 will itself lead to a remainder 1....

hence R = 1

A = 1 + 1^2 ....... 1^n A = n

$i$ $did$ $it$ $like$ $this:$ $2^3 \equiv 1 \pmod{7}$

$\Rightarrow$ $2^{999} \equiv 1 \pmod{7}$

$thus$ $as$ $we$ $see,$ $we$ $get$ $1$ $as$ $remainder....$

$hence,$

$A=1+1+1..... n times = 1*n = n$

Remainder will be 1 so the value of A will be n.