1. In the set of natural numbers, solve the equation

$$
x^{5-x}=(6-x)^{1-x}
$$
Solution. Direct verification shows that among the numbers $1,2, \ldots, 6$, only 1 and 5 satisfy the given equation. Let us assume that some natural number $x>6$ is a solution to the equation. Clearly, $x$ is an odd number, i.e., $x=2 k+1, k \geq 3$. Substituting into the equation, we obtain the equation $(2 k+1)^{4-2 k}=(5-2 k)^{-2 k}$, which is equivalent to the equation

$$
\left(1+\frac{6}{2 k-5}\right)^{k-2}=(2 k-5)^{2}
$$

Accordingly, $\frac{6}{2 k-5}$ must be an integer, i.e., $2 k-5 \in\{ \pm 1, \pm 2, \pm 3, \pm 6\}$. The values $\pm 2$ and $\pm 6$ are even, so they are excluded, and for the values -1 and -3, we get $2 k+1<7$. Only $2 k+1=7$ and $2 k+1=9$ remain. Direct verification shows that in the first case, there is no solution, and in the second case, we have the solution $x=9$.

Finally, the solutions to the given equation are $x=1, x=5, x=9$.