ANY LAST WORDS, LATEX BRAGGER?

Show that from any five integers, one can always choose three of these integers such that their sum is divisible by 3.


The possible residues of modulo 3 are 0,1, and 2.

For this to be divisible by 3, the sum should be modulo 0.

Let:
$$A\equiv 0\pmod3$$
$$B\equiv 1\pmod3$$
$$C\equiv 2\pmod3$$

Let's see what happens if any of $A, B, C$ are tripled.
$$(0\pmod3)+(0\pmod3)+(0\pmod3)\equiv (0\pmod3)$$
$$(1\pmod3)+(1\pmod3)+(1\pmod3)\equiv (3\pmod3)\equiv (0\pmod3)$$
$$(2\pmod3)+(2\pmod3)+(2\pmod3)\equiv (6\pmod3)\equiv (0\pmod3)$$
Since there are three of the same term, we are basically multiplying by $3$ which is another way to calculate whether or not it is divisible by 3


So as long as there are three of $A, B, C$, the other two terms won't matter, because we will always pick those three.

Now it's time to list out the other three possibilities.
$$\begin{array}{|c|c|c|c|c|} \hline A& A & B & B &C\\ \hline A&A&B&C&C\\ \hline A&B&B&C&C\\ \end{array}$$


We begin by noticing that $A+B+C=0+1+2=3=0$, and the three possibilities each have an $A, B, C$. Therefore, from any five integers, one can always choose three of these integers such that their sum is divisible by 3