1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ What Kind of numbers are $x,y,z$? $\endgroup$ Apr 12, 2018 at 13:59
  • $\begingroup$ Any condition on $z$? $\endgroup$ Apr 12, 2018 at 14:06
  • $\begingroup$ I suppose, $x,y,z\in \mathbb{Z}$. $\endgroup$ Apr 12, 2018 at 14:14
  • $\begingroup$ No conditions were given for z $\endgroup$
    – Janjan
    Apr 12, 2018 at 14:35

5 Answers 5

5
$\begingroup$

Since $2018=2\cdot 1009$, we have only the divisors $1,2,1009,2018$, so we have $$ r_2(2018)=4(d_1(n)-d_3(n))=4, $$ where $$d_i(n)=\sum_{d\mid n,d\equiv i \bmod 4}1.$$ So there are only $4$ possibilities, up to sign and permutation, namely essentially only $$ 13^2+43^2=2018. $$ Then we solve $xyz=11(x+y+z)$ for all $4$ possibilities $(x,y)=(\pm 13,\pm 43)$. This is not solvable for integer $z$.

$\endgroup$
3
$\begingroup$

First $x^2+y^2=2018$ has as only solutions $(x,y)=(\pm13,\pm43),(\pm43,\pm13)$ so by the restriction $x\le y$ we have two solutions $(x,y)=(\pm13,43)$. This is not compatible with the equation $xyz=11(x+y+z)$ because in both cases we would have $$-z=\frac{330}{570}\\z=\frac{616}{548}$$ Thus there are no solutions.

$\endgroup$
2
$\begingroup$

For real numbers, we have $$z=\frac{11x+11y}{xy-11}=\frac{22(x+y)}{(x+y)^2-2040}$$ subject to $-2\sqrt{1009}\leq x+y\leq 2\sqrt{1009}$

$\endgroup$
2
$\begingroup$

To use as little machinery as possible but at the cost of a bit of computation, note that we must have $\frac 12\cdot 2018 \lt y^2 \lt 2018$, which gives $33 \le y \le 44$, only $12$ possibilities. Trying each to check if $x$ is an integer gets us that the only choice for $x,y$ is $x=13, y=43$. That would require $13\cdot 43z=11(13+43+z)$ which clearly has no solution with $z \ge 1$ because the left side is much too big.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Why can't $z$ be the one divisible by $11$? $\endgroup$ Apr 12, 2018 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.