The end digits of those 8 consecutive numbers can be from 1 to 8 or from 2 to 9.

We can write that sum of 8 consecutive numbers as:

(for 1-8)

(k10^n + 1) + (k10^n + 2) + (k10^n + 3) + (k10^n + 4) + (k10^n + 5) + (k10^n + 6) + (k10^n + 7) + (k10^n + 8) =

k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 =

k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + 36

or as (for 2-9)

k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + 44

As I understand 8 consecutive numbers must be greater then 10.

Since in 8 consecutive numbers there is always 2 as last digit then k10^n is divisible by 2 for k>0 and n>0 and so is 36 (or 44) divisible by 2 so we get whole number.

Similar can be done for b) and c).