1. (AUS 6) Let $f$ be a function that satisfies the following conditions: (i) If $x>y$ and $f(y)-y \geq v \geq f(x)-x$, then $f(z)=v+z$, for some number $z$ between $x$ and $y$. (ii) The equation $f(x)=0$ has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions; (iii) $f(0)=1$. (iv) $f(1987) \leq 1988$. (v) $f(x) f(y)=f(x f(y)+y f(x)-x y)$.  Find $f(1987)$.
1. By (ii), $f(x)=0$ has at least one solution, and there is the greatest among them, say $x_{0}$. Then by (v), for any $x$, 
$$ 
0=f(x) f\left(x_{0}\right)=f\left(x f\left(x_{0}\right)+x_{0} f(x)-x_{0} x\right)=f\left(x_{0}(f(x)-x)\right) 
$$ 
It follows that $x_{0} \geq x_{0}(f(x)-x)$. Suppose $x_{0}>0$. By (i) and (iii), since $f\left(x_{0}\right)-x_{0}<0<f(0)-0$, there is a number $z$ between 0 and $x_{0}$ such that $f(z)=z$. By (1), $0=f\left(x_{0}(f(z)-z)\right)=f(0)=1$, a contradiction. Hence, $x_{0}<0$. Now the inequality $x_{0} \geq x_{0}(f(x)-x)$ gives $f(x)-x \geq 1$ for all $x$; so, $f(1987) \geq 1988$. Therefore $f(1987)=1988$.