2. The following equation is to be solved in the set of natural numbers

$$
2^{x}+2^{y}+2^{z}=2336
$$
Solution. If an ordered triplet of natural numbers is a solution to the equation, then any of its permutations is also a solution to the equation.

Let $x, y, z$ be a solution to the equation in the set of natural numbers. We will show that $x \neq y \neq z \neq x$, by showing that if any two numbers from $x, y, z$ are equal, then $x, y, z$ is not a solution. Let $x=y$. Then it is possible: 1) $z=x$; 2) $z=x+1$.

1) If $z=x$, then $2^{x}+2^{y}+2^{z}=3 \cdot 2^{x}$ is divisible by 3, while 2336 is not divisible by 3.
2) If $z=x+1$, then 2336 is divisible by 73, while $2^{x}+2^{y}+2^{z}=2^{x+2}$ is not.
3) Let $z>x+1$, i.e., $z=x+1+k$ for some natural number $k$. Then

$$
2^{x}+2^{y}+2^{z}=2^{x+1}\left(1+2^{k}\right)
$$

is not equal to 2336 for the same reasons as in 2).

Similarly, it can be shown that $x \neq z$ and $y \neq z$.

Let $x<y<z$, i.e., $y=x+p, z=y+s=x+p+s$. Then from

$$
2^{x}\left[1+2^{p}\left(1+2^{s}\right)\right]=2^{x}+2^{y}+2^{z}=2336=2^{5}\left[1+2^{3}\left(1+2^{3}\right)\right]
$$

it follows that $x=5, y=8$ and $z=11$. Therefore, the solutions to the equation are all permutations of the triplet $5,8,11$.