Guessing Won't Work!!!!!

Every prime can be expressed as

6k±1 except for 2 and 3.

Thus,

6k±1 $\equiv\ ±1 (mod 6 )$

⇒ $(6k±1)^{2}$ $\equiv\ 1 (mod 6 )$

Now, we are left with 2 and 3,

$2^{2}$ $\equiv\ 4 (mod 6 )$

$3^{2}$ $\equiv\ 3 (mod 6 )$

Thus,

$S_{r}$ = m =1+4+3= $\boxed{8}$

we know that 1000(n) is always divisible by 8 for any value of n as an integer..

1000(100)+1 $\equiv\ 1 (mod 8 )$

⇒ $(100001)^{100001}$ $\equiv\ 1 (mod 8 )$

Thus,

we get the answer $\huge{1}$