<s>[INST] $xy=22$ and $yz=26$: What is $x+y+z $ equal to? Given the following: $$xy=22,\qquad yz=26,$$ where $x,y,z\in\mathbb{N}$. Which of the following is a possible value of $ x + y + z $? 
$ \textbf {(A) } 22 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 26 \qquad \textbf {(D) } 48 $ 
 [/INST] Look at $xy$ divisors  and $yz$ divisors. (Because we have that $x,y,z \in \mathbb{N}$.)
$$
\begin{array}{|c|lcr|}
\hline xy \text{ and } yz & \text{} \\
\hline
22 & 1 & 2 & 11 & 22 \\
26 & 1 & 2 & 13 & 26 \\
\hline
\end{array}
$$
So if $xy=22=\color{red}{2}\cdot11 = \color{green}{1} \cdot 22$, and $yz=26=\color{red}{2}\cdot13 = \color{green}{1} \cdot 26$ 
then $y$ is either $\color{red}{2}$ or $\color{green}{1}$ and nothing else. We can do an easy shortcut: In the possibilities, we only have even numbers. $(1)$ 
So $y$ cannot be equal to $1$ since it would imply that $x+y+z$ is odd in contradiction with $(1)$. 
But $x+y+z$ will be even if $y = 2$, so: $x+y+z=11+2+13=26$
Therefore the right answer is: $\boxed{\textbf{(C) } 26}$
</s>