Gukhanov $H . X$.

Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. From each trinomial $f_{i}(x)$, one root was chosen and denoted as $x_{i}$. What values can the sum $f_{2}\left(x_{1}\right)+$ $f_{3}\left(x_{2}\right)+\ldots+f_{100}\left(x_{99}\right)+f_{1}\left(x_{100}\right)$ take?
Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then $f_{2}\left(x_{1}\right)=f_{1}\left(x_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$, since $f_{1}\left(x_{1}\right)=$ 0. Similarly, we obtain the equalities

$f_{3}\left(x_{2}\right)=c_{3}-c_{2}, \ldots, f_{100}\left(x_{99}\right)=c_{100}-c_{99}$ and $f_{1}\left(x_{100}\right)=c_{1}-c_{100}$.

By adding, we get $f_{2}\left(x_{1}\right)+f_{3}\left(x_{2}\right)+\ldots+f_{1}\left(x_{100}\right)=\left(c_{2}-c_{1}\right)+\ldots+\left(c_{1}-c_{100}\right)=0$.

## Answer

Only 0.