The integers $x, y$ and $z$ satisfy $\frac{x}{5}=\frac{6}{y}=\frac{z}{2}$. What is the largest possible value of $x+y+z ?$
Since $\frac{x}{5}=\frac{6}{y}$, then $x y=30$.

Since $\frac{6}{y}=\frac{z}{2}$, then $y z=12$.

Since we would like the maximum value of $x+y+z$, we may assume that each of $x, y$ and $z$ is positive.

Since $x y=30$ and $y z=12$ and $y$ is a positive integer, then $y$ is a divisor of each of 30 and 12 . This means that $y$ must equal one of $1,2,3$, or 6 .

If $y=1$, then $x=30$ and $z=12$ which gives $x+y+z=43$.

If $y=2$, then $x=15$ and $z=6$ which gives $x+y+z=23$.

If $y=3$, then $x=10$ and $z=4$ which gives $x+y+z=17$.

If $y=6$, then $x=5$ and $z=2$ which gives $x+y+z=13$.

Therefore, the maximum possible value of $x+y+z$ is 43 .

ANSWER: 43