Zlobin C.A.

Two different numbers $x$ and $y$ (not necessarily integers) are such that $x^{2}-2000 x=y^{2}-2000 y$. Find the sum of the numbers $x$ and $y$.
We can rewrite the given equation as follows: $2000(x-y)=(x-y)(x+y)$. From this, $x+y=2000$ (since $x \neq y$, we can cancel out $x-y$).

## Answer

2000.