1015*. Find a three-digit number such that its product by 6 is a three-digit number with the same sum of digits. List all such solutions.
$\triangle$ Let the desired number be denoted as $\overline{a b c}$, and the new number as $\overline{\mathrm{klm}}$.

Then

$$
\overline{a b c} \cdot 6=\overline{k l m}
$$

where, by the condition,

$$
a+b+c=k+l+m
$$

From this, it is clear that $\overline{a b c} \leqslant 166$, since the product $167 \cdot 6=1002$.

The number $\overline{k l m}$ is divisible by 3, i.e., by the divisibility rule for 3, the sum $k+l+m$ is divisible by 3. Therefore, the sum $a+b+c$ is divisible by 3, which means the number $\overline{a b c}$ is divisible by 3. But in this case, the number $\overline{k l m}$ is already divisible by 9. Then

$$
(k+l+m): 9, \quad(a+b+c): 9, \quad \overline{a b c}: 9
$$

Let's find all values of the number $\overline{a b c}=\overline{1 b c}$, remembering that it does not exceed 166:

$$
108,117,126,135,144,153,162
$$

The number $\overline{k l m}$ is 6 times larger, so it takes the following values accordingly:

$$
648,702,756,810,864,918,972.
$$

Now let's compare the sums of the digits of the numbers in the first row with the sums of the digits of the corresponding numbers in the lower row. When are they equal? Only in two cases: when $\overline{a b c}=117$ and when $\overline{a b c}=135$.

Answer: $117,135$.