## Task B-1.6.

The digit in the tens place of a three-digit number is 9, and the sum of its digits is divisible by 6. If we swap the digits in the hundreds and units places of this number, the new number is 99 less than twice the original number. Determine the original three-digit number.
## First Solution.

We write the three-digit number in the form $\overline{a 9 b}$. Since $a$ and $b$ are digits different from 0 and the sum of the digits is divisible by 6, and less than 27, we have

$$
a+9+b \in\{12,18,24\} \text {, i.e., } a+b \in\{3,9,15\}
$$

Furthermore, according to the condition in the problem, we have $\overline{b 9 a}=2 \cdot \overline{a 9 b}-99$, i.e.,

$$
\begin{array}{rlrl}
100 b+90+a & =2(100 a+90+b)-99 . & 1 \text { point } \\
199 a-98 b & =9 . & \text { (**) } & 1 \text { point }
\end{array}
$$

According to $(\star)$, we have three possibilities:

$$
b=3-a \text { or } b=9-a \text { or } b=15-a \text {. }
$$

Let's check all three by substituting into ( $\star \star$ ).

If $b=3-a$, then we have:

$$
\begin{aligned}
199 a-98(3-a) & =9 \\
199 a-294+98 a & =9 \\
297 a & =303 .
\end{aligned}
$$

This equation clearly has no integer solution.

If $b=9-a$, then we have:

$$
\begin{aligned}
199 a-98(9-a) & =9 \\
199 a-882+98 a & =9 \\
297 a & =891 \\
\Longrightarrow a=3, b & =9-3=6
\end{aligned}
$$

If $b=15-a$, then we have:

$$
\begin{aligned}
199 a-98(15-a) & =9 \\
199 a-1470+98 a & =9 \\
297 a & =1479
\end{aligned}
$$

The last equation has no integer solution because the left side is divisible by 9, but the right side is not.

Therefore, the desired three-digit number is 396.