Possible values of $x+y+z$ Given that $xyz=11(x+y+z)$, find the possible values of $x+y+z$ such that $x\leq y$ satisifies $x^2+y^2=2018$
I know that $x,y$, or $z$ should be divisible by $11$. I am having trouble going to the next step

I think the problem also states that $x,y,z$ must be integers.
Assume that $x$ is divisible by $11$, then $x=11k$ ($k$ is an integer) and $(11k)^2+y^2=2018$.
We will have $y^2=2018-121k^2$. Because $y^2\ge 0$, we can prove that $k^2$ must be in this set of numbers: $(1;4;9;16)$.
We have four cases:


*

*$k^2=1\Rightarrow y=43.55$, result eliminated

*$k^2=4\Rightarrow y=39.17$, result eliminated

*$k^2=9\Rightarrow y=30.48$, result eliminated

*$k^2=16\Rightarrow y=9.05$, result eliminated
So there are no cases available for "$x$ is divisible by $11$" that satisfy $x^2+y^2=2018$, this means "$y$ is divisible by $11$" cannot be true as well.
Because of this, someone should attempt it by solving the equation $x^2+y^2=2018$ for integers $x,y$ instead, you can make a table that consists of $45$ cases from $1$ to $45$ for $x$ to find $y$ (or fewer cases if you use the condition $x\le y$ to eliminate some cases). This method is really tedious, but you should get $x=13$ and $y=43$, then you can solve for $z$.
However, $z=1.12$ which means $z$ is not an integer, this means there are no three integers $x,y,z$ satisfy all the conditions.
