Let $[x]$ be the integer part of a number $x$, and $\{x\}=x-[x]$. Solve the equation

$$
[x] \cdot\{x\}=1991 x .
$$
Let $f(x)=[x] \cdot\{x\}$. Then we have to solve the equation $f(x)=1991 x$. Obviously, $x=0$ is a solution. For any $x>0$ we have $0 \leq[x] \leq x$ and $0 \leq\{x\}[x]>x-1$ and $0 \leq\{x\}x-1>1991 x$. Finally, if $-1<x<0$, then $[x]=-1,\{x\}=x-[x]=x+1$ and $f(x)=-x-1$. The only solution of the equation $-x-1=1991 x$ is $x=-\frac{1}{1992}$.