5. Let $x, y, z$ be positive integers such that
$$
\begin{array}{l}
(x+y)(y+z)=2016 \\
(x+y)(z+x)=1080
\end{array}
$$

Determine the smallest possible value for $x+y+z$.
Answer: 61
Solution: Note that $2016=2^{5} \times 3^{2} \times 7$ and $1080=2^{3} \times 3^{3} \times 5$. Moreover
$$
x+y+z=\frac{1}{2}((x+y)+(y+z)+(z+x))
$$

Since $x+y$ is a common factor for both 2016 and 1080, and we want $x+y+z$ to be as small as possible, then we try to find the largest possible factor for 2016 and 1080 such that $x, y, z$ are integers
$$
\begin{array}{l}
(x+y)(y+z)=\left(2^{3} \times 3^{2}\right) \times\left(2^{2} \times 7\right)=72 \times 28 \\
(x+y)(z+x)=\left(2^{3} \times 3^{2}\right) \times(3 \times 5)=72 \times 15
\end{array}
$$
hence $x+y=72, y+z=28, z+x=15$. But $x+y+z=\frac{1}{2}(72+28+15)=57.5$ which cannot be since $x, y, z$ are integers.
Therefore we try the following:
$$
\begin{array}{l}
(x+y)(y+z)=\left(2^{2} \times 3^{2}\right) \times\left(2^{3} \times 7\right)=36 \times 56 \\
(x+y)(z+x)=\left(2^{2} \times 3^{2}\right) \times(2 \times 3 \times 5)=36 \times 30
\end{array}
$$
hence $x+y+z=\frac{1}{2}(36+56+30)=61$. This is the smallest possible sum given that $x, y, z$ are positive integers.