$xy=22$ and $yz=26$: What is $x+y+z $ equal to?

Question

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 $

Answer 1

Hint: There aren't very many choices for $x, y, z$. In fact, since $22 = 2 \cdot 11$ and $22 = 1 \cdot 22$ are the only ways to break up $22$ as multiplication, we see that

$$x, y \in \{1, 2, 11, 22\}$$

On the other hand,

$$y, z \in \{1, 2, 13, 26\}$$

Combining these shows that $y \in \{1, 2\}$. Now consider cases based on this.

Answer 2

Shortcut. Since both given products are even, we can just assume that $y$ is even, so let it be $y=2$. Then, $x=11$ and $z=13$. Then, $$x+y+z=11+2+13=\boxed{\textbf{(C) } 26}.$$

Answer 3

We know that $(x+z)y=48$. We are looking for $(x+z)+y$. $$ x+z=1;y=48\implies (x+z)+y=49\\ x+z=2;y=24\implies (x+z)+y=26\\ x+z=3;y=16\implies (x+z)+y=19\\ x+z=4;y=12\implies (x+z)+y=16\\ x+z=6;y=8\implies (x+z)+y=14\\ $$ we could switch the values, and get the other choices, but the sums and products remain the same.

The only common element is $26$.

If we want the actual values, note that $(x+z)y=2\cdot24$. Since neither $xy=22$ nor $yz=26$ are divisible by $24$, we need $y=2$. Then we get $x=11$ and $z=13$.

Answer 4

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: $\boxed{\textbf{(C) } 26}$