595*. Does there exist a three-digit number whose cube ends in three sevens?
Let's denote this three-digit number as $x$. Then

$$
x^{3}=1000 a+777
$$

where $a$ is the number of thousands in $x^{3}$.

Since $x^{3}$ ends with the digit 7, the number $x$ ends with the digit 3: $x=10 k+3$, and the number $k$ is two-digit. We obtain:

$$
\begin{gathered}
(10 k+3)^{3}=1000 a+777, \quad 1000 k^{3}+900 k^{2}+270 k+27=1000 a+777 \\
100 k^{3}+90 k^{2}+27 k=100 a+75
\end{gathered}
$$

From this, it is clear that the number $k$ is divisible by 5.

Since then, in the left part of the last equation, the terms $100 k^{3}$ and $90 k^{2}$ are divisible by 25, and the right part of the equation is also divisible by 25, $k$ is divisible by 25: $k=25 b$, where $b$ is a digit not exceeding 3. We obtain:

$$
x=10 k+3=250 b+3
$$

Let's consider three possibilities.

1) Suppose $b=1$. Then:

$$
x=253, \quad x^{3}=253^{3}=16194277
$$

This case does not fit, as the number $x^{3}$ ends with two, not three sevens.

2) Suppose $b=2$. Therefore,

$$
x=503, \quad x^{3}=503^{3}=117263527
$$

This case also does not fit.

3) Suppose $b=3$. We get:

$$
x=753, \quad x^{3}=753^{3}=426957777
$$

This variant satisfies the condition of the problem: the number $x^{3}$ ends not with three, but with four sevens.

Answer: It exists and is unique - 753.