Suppose x, y, z are three positive integers such that x < yz and xyz = Which one of the following values of S yields more than one solution to the equation x+y+z=S?

The correct answer is (b).

Suppose $x, y, z$ are three positive integers such that $x < yz$ and $xyz = 13$. We can factor $13$ as $13 = 1 \cdot 13 = 3 \cdot 4 = 6 \cdot 2$. Therefore, we can choose $x, y, z$ to be any of the following ordered triples:

• $(1, 13)$
• $(3, 4)$
• $(6, 2)$

Since there are three possible solutions, $S = 13$ yields more than one solution to the equation $x + y + z = S$.

For the other options, we can see that there is only one solution to the equation $x + y + z = S$ for each value of $S$.

• For $S = 4$, the only solution is $(1, 1, 2)$.
• For $S = 15$, the only solution is $(3, 5, 3)$.
• For $S = 16$, the only solution is $(4, 4, 4)$.