Determine the number of these subsets for which the sum of the three elements is a multiple of 3. [Read Full Q. Below]

Question

Consider the collection of all three-element subsets drawn from the set ${1,2,3,4,..., 299, 300}$. > Determine the number of these subsets for which the sum of the three elements is a multiple of 3.

This is a question from the Chapter Combinatorics but from what I understood of the question, I tried to find a pattern between subsets from which I found that the sum of all consecutive 3 positive integers is a multiple of 3, meaning if repetition isn't allowed, the three-element subsets would be $100$

I can't find a solution if repetition is allowed so please verify my answer if repetition isn't allowed and answer the question thinking it is allowed.

EDIT/UPDATE: IF repetition is allowed and we take the fact that the sum of all 3 consecutive integers is a multiple of 3 then each consecutive set {1, 2, 3}, {4, 5, 6}, will have 3 more sets for example the first will also be viable for {2, 3, 4}, {3, 4, 5}; also, a set with the same number repeating 3 times would also be a multiple of 3 so the final answer would be $3 * 100 + 300 = 600$ Would this be correct?

Answer 1

I'm considering the problem on top of your question. There is no "consecutive" involved, and "repetition" does not occur in sets.

We have $x_1+x_2+x_3=0$ modulo $3$ iff either all $x_i$ have different remainders modulo $3$, or all $x_i$ have the same remainder modulo $3$. It follows that there are $${100\choose1}^3+3{100\choose3}=1\,485\,100$$ three-element subsets of $[300]$ whose sum is divisible by $3$.

Answer 2

For your $S=\{1,2,3,\ldots, 300\}$ write $S=A\cup B\cup C$ with $$A=\{3,6,9,\ldots, 300\}=S\cap 3\Bbb Z $$ $$B=\{1,4,7,\ldots, 298\}=S\cap(3\Bbb Z+1) $$ $$C=\{2,5,8,\ldots, 299\}=S\cap(3\Bbb Z+2) $$

There are $300\choose 3$ 3-subsets of $S$. Subtract the $100\choose 3$ sets that are $\subset A$. Subtract the $100\choose 3$ sets that are $\subset B$. Subtract the $100\choose 3$ sets that are $\subset C$. Subtract the $100^3$ sets that correspond to triples in $A\times B\times C$.