A three-digit number composed of different digits, 75% of which is made up of the same digits as the original, but none of them remain in their original position. Which number is this?
Solution. There are two cases:

I. case: It might be that $\frac{3}{4} \cdot \overline{a b c}=\overline{b c a}$, then

$$
\frac{3}{4}(100 a+10 b+c)=100 b+10 c+a
$$

which means

$$
\begin{aligned}
300 a+30 b+3 c & =400 b+40 c+4 a \\
296 a & =37 c+370 b \\
8 a & =c+10 b
\end{aligned}
$$

That is, $8 a=\overline{b c}$, which is an 8-divisible number, at most two digits.

| $a$ | $b$ | $c$ | Original number | New number |
| :---: | :---: | :---: | :---: | :---: |
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 8 | 108 | 081 |
| 2 | 1 | 6 | $\mathbf{2 1 6}$ | 162 |
| 3 | 2 | 4 | $\mathbf{3 2 4}$ | 243 |
| 4 | 3 | 2 | $\mathbf{4 3 2}$ | 324 |
| 5 | 4 | 0 | $\mathbf{5 4 0}$ | 405 |
| 6 | 4 | 8 | $\mathbf{6 4 8}$ | 486 |
| 7 | 5 | 6 | $\mathbf{7 5 6}$ | 567 |
| 8 | 6 | 4 | $\mathbf{8 6 4}$ | 648 |
| 9 | 7 | 2 | $\mathbf{9 7 2}$ | 729 |

The three-digit numbers highlighted in dark are the solutions to the problem. (108 is not a solution, as its 75% is 81, and this number does not contain a zero.)

II. case: It might be that $\frac{3}{4} \cdot \overline{a b c}=\overline{c a b}$, then

$$
\frac{3}{4}(100 a+10 b+c)=100 c+10 a+b
$$

which means

$$
300 a+30 b+3 c=400 c+40 a+4 b, \quad 260 a+26 b=397 c
$$

The expression on the left side is divisible by 13, while the one on the right side is not. (c is a positive single-digit number.) Thus, in this case, we do not get a solution.