Let's determine the distinct digits $A, B, C$ if the product of the number " $\overline{A B}$ whole, $C$ tenth" and $C$, rounded to the nearest integer, results in the number $\overline{B C}$: $\overline{A B, C} \cdot C \approx \overline{B C}$.
If $A, B, C$ are distinct, then it is obvious that $C$ cannot be 1. $B$ must be at least $A \cdot C$, because there must be at least this many tens in the product, so $A$ and $C$ must be larger than $A$. If $C \geq 4$, then $A$ can only be 1, because otherwise the product would be greater than $25 \cdot 4=100$. It cannot be $C \geq 6$, because then the product would be greater than $17 \cdot 6>100$. Therefore, we need to examine the cases $C=2,3,4$, and 5.

We can write our approximate equality in the form of the following double inequality (allowing both rounding up and down of the tens digit):

$$
\overline{B C}-0.5 \leq \overline{A B, C} \cdot C \leq \overline{B C}+0.5
$$

Multiplying by 10 and subtracting $\overline{B C} \cdot C$:

$$
(10-C) \cdot \overline{B C}-5 \leq 100 \cdot A \cdot C \leq (10-C) \cdot \overline{B C}+5
$$

Adding 5 to the first inequality and subtracting 5 from the second, we can then recombine the resulting two inequalities in reverse order:

$$
100 \cdot A \cdot C-5 \leq (10-C) \cdot \overline{B C} \leq 100 \cdot A \cdot C+5
$$

Thus, $(10-C) \cdot \overline{B C}$, when rounded up or down to the nearest ten, must result in a number divisible by 100, where the number of hundreds is divisible by $C$. This is impossible for $C=5$, because then the number of tens in the middle product would be 2 or 7. If $C=4$, then on one hand $A=1$, and on the other hand, the middle $6 \cdot \overline{B 4}$ must be rounded down, so $6 \cdot B$ must end in 8, meaning $B=3$ or 8. The former is smaller for smaller $C$, and with the latter, rounding gives 500 instead of 400.

If $C=3$, the middle $7 \cdot \overline{B 3}$ will be rounded to a multiple of 100 when $B=4$, and indeed 300, which is of the form $100 \cdot A \cdot C$ for $A=1$. Indeed, $14.3 \cdot 3=42.9 \approx 43$, meeting the requirements of the problem.

If $C=2$, $8 \cdot \overline{B 2}$, when rounded to the nearest ten (upward), only gives round hundreds for $B=1$ or 6. The former is smaller for smaller $C$, and with the latter, the rounded product is 500, and the number of hundreds is not divisible by $C$. Thus, the only solution to the problem is $A=1, B=4, C=3$.