593*. Find some power of the number 2 with a natural exponent that ends in three identical digits.
Let's denote the number of thousands in the number $2^{k} (k \in \mathbb{N})$ by $x$, and the digit that appears three times at the end of this number by $y$. Then,

$$
2^{k}=1000 x+\overline{y y y}=1000 x+111 y
$$

From this, it follows that $2^{k}>1000$, and thus $k \geqslant 10$. We obtain that in this equation, $1000 x$ is divisible by 8 and $2^{k}$ is divisible by 8, which means:

$$
111 y \div 8, \quad y \div 8 \Rightarrow y=8
$$

Our equation becomes:

$$
2^{k}=1000 x+888, \quad 2^{k-3}=125 x+111,
$$

where the number $x$ is odd.

When $x$ is odd, the number $125 x$ ends in 125, 375, 625, or 875 (verify this!), and thus the sum $125 x+111$ ends in

$$
236, 486, 736, 986.
$$

Now the problem reduces to finding at least one natural $k$ for which the power $2^{k-3}$, where $k \geqslant 10$, ends in one of the four last three-digit numbers.

Since the power $2^{k-3}$ ends in 6, the exponent is divisible by 4:

$$
k-3=4 a, \quad k=4 a+3
$$

where $a$ is a natural number not less than 2.

Let's apply a brute-force search for $a$.

1) Let $a=2$. Then

$$
k-3=4 \cdot 2=8, \quad 2^{k-3}=2^{8}=256
$$

But the number 256 is not among the above three-digit numbers.

2) For $a=3$

$$
k-3=12, \quad 2^{k-3}=2^{12}=4096
$$

but this option also does not work.

A similar situation occurs for $a=4, 5, 6, 7,$ and 8.

3) Let $a=9$. Then

$$
k-3=36, \quad k=39, \quad 2^{k-3}=2^{36}
$$

and $2^{36}$ ends in 736 (think about how to establish this). The number 736 is among the four three-digit numbers, so this case works.

Answer: for example, $2^{39}$ ends in 888.