hello before i post these for any mods or whatever, i'm kind of unclear on whether it's allowed to post problems and things like that (this isn't my homework). Please be easy with the rules lol, new here.

Ok so i may be a bit of a retard but im having a hard time with part (b). Could anyone nudge me? I will post my solution to part (a). probably missing something obvious :d

(a) Prove that given any $5$ integers, there exist three of them whose sum is divisible by 3.
Let S be the list of integers (a_1, a_2, a_3, a_4, a_5). For 1 <= i <= 5, let z_i belong to {0,1,2}, z_i = a_i mod 3. We need to prove that for every instance of the list Z = (z_1, z_2, z_3, z_4, z_5) it contains three numbers z_i, z_j, z_k with a sum 0, 3 or 6; in which case the corresponding integers a_i, a_j, a_k of the list S would have a sum divisible by 3.

We proceed by contradiction and assume that no three elements of the list Z add to 0, 3 or 6. Observe that in this case Z cannot contain 0, 1 and 2 at the same time since 0 + 1 + 2 = 3. Without loss of generality, we assume that 0 does not occur in the list and it contains only 1s and 2s. By the pigeonhole principle then we see that the list has either three 1s or three 2s, which gives us a contradiction, since 1 + 1 + 1= 3 and 2 + 2 + 2= 6 .

and now what i am stumped on :
(b) Prove that given any 17 integers, there exist nine of them whose sum is divisible by 9.