Find The Largest X

We must have $a_j \ge 0$ . Furthermore, the values of $a_j$ must be distinct. Therefore, if we rearrange the values of $a_j$ in order from least to greatest (let's call the new sequence $a_{m_j}$ ), then we have $a_{m_1} \ge 0, a_{m_2} \ge 1, a_{m_3} \ge 2$ and so on. In general $a_{m_j} \ge j - 1$ . Therefore, $\sum_{j = 1}^{x} a_j = \sum_{j = 1}^{x} a_{m_j} \ge \sum_{j = 1}^{x} (j-1) = \frac{(x-1)x}{2}$ Similarly, $\sum_{j = 1}^{x} b_j \ge \frac{(x-1)x}{2}$ and $\sum_{j = 1}^{x} c_j \ge \frac{(x-1)x}{2}$ Adding these three inequalities gives:

$\frac{3(x-1)x}{2} \le \sum_{j = 1}^{x} (a_j+b_j+c_j) = 2014x$

$x \le \frac{2}{3} *2014 + 1$ Therefore $X \le 1343$ .

Now let's show that it is possible to create 1343 such triples. We create them as follows. Suppose $1 \le j \le 1343$ .

If $j\ = 2m - 1$ for some $m \in \mathbb{Z}$ , let $(a_j, b_j, c_j) = (m - 1, 670 + m, 1345 - 2m)$ ;

If $j\ = 2m$ for some $m \in \mathbb{Z}$ , let $(a_j, b_j, c_j) = (671 + m , m - 1, 1344 - 2m)$ .

It's not difficult to show that the values of $a_j$ are nonnegative and pairwise distinct, and likewise with $b_j$ and $c_j$ . Also, for each $j$ , we have $a_j + b_j + c_j = 2014$ . Hence this list of triples satisfies the conditions.

Therefore, it is possible to create 1343 triples, and we showed earlier that the number of triples cannot exceed 1343.

Therefore, $X = 1343$ .