South African Mathematics Olympiad

Suppose the numbers are $x$, $y$, $z$, with $0 < x < y < z$ and $x + y + z = 7$. If $x \ge 2$, then $y \ge 3$ and $z \ge 4$, so $x + y + z \ge 9$, which is too large. Therefore $x = 1$, which leaves $y + z = 6$. Since $2 \le y < z$, the only possibility is $y = 2$ and $z = 4$. The product $xyz = 1 \times 2 \times 4 = 8$.