Balti Tee 1991 · Ülesanne 6

Solution: 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 $0 \leq \{x\} < 1$ which imply $f(x) < x < 1991 x$.

For $x \leq -1$ we have $0 > [x] > x - 1$ and $0 \leq \{x\} < 1$ which imply $f(x) > x - 1 > 1991 x$.

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 $x = -\frac{1}{1992}$.