1067. Kolya had to square a certain natural number for his homework. Instead, he mistakenly doubled the number and got a two-digit number, written with the same digits as the square of the number, but in reverse order. What should the correct answer be?
$\triangle$ Let's denote the original number by $x$, its square - by $\overline{a b}$. Then

$$
\left\{\begin{array}{l}
x^{2}=10 a+b \\
2 x=10 b+a
\end{array}\right.
$$

From here, it is clear that the number $x$ is a single-digit number.

Let's add the equations of this system:

$$
x(x+2)=11(a+b)
$$

Then either $x$, or $x+2$ equals 11. But the equality $x=11$ is impossible, and when $x+2=11$ we find:

$$
x=9, \quad \overline{a b}=81
$$

We need to check the second equation of the system. For $x=9, a=8, b=1$ the equality holds.

Answer: 81.