10.1. 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. For each trinomial $f_{i}(x)$, one root was chosen and denoted by $x_{i}$. 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?
Answer: Only 0.

Solution. Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then

$$
f_{2}\left(x_{1}\right)=a x_{1}^{2}+b x_{1}+c_{2}=\left(a x_{1}^{2}+b x_{1}+c_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1},
$$

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 $f_{1}\left(x_{100}\right)=c_{1}-c_{100}$.

Adding the obtained equalities, we get $f_{2}\left(x_{1}\right)+f_{3}\left(x_{2}\right)+\ldots+f_{1}\left(x_{100}\right)=\left(c_{2}-c_{1}\right)+\ldots+\left(c_{1}-c_{100}\right)=0$. Therefore, the only possible value of the sum is zero.

Comment. A correct answer without justification - 0 points.