B1. Find all natural numbers $n$ of the form $n=\overline{23 a b 16 c}$, which have all distinct digits and are divisible by 9 and 11. Here, $a, b$, and $c$ are digits.
B1. The number $n$ is divisible by 99. We can write

$$
\begin{aligned}
n & =\overline{23 a b 1} \cdot 100+60+c=\overline{23 a b 1} \cdot 99+\overline{23 a b 1}+60+c= \\
& =\overline{23 a b 1} \cdot 99+\overline{23 a} \cdot 100+10 b+1+60+c= \\
& =(\overline{23 a b 1} \cdot 99+\overline{23 a}) \cdot 99+\overline{23 a}+10 b+c+61= \\
& =(\overline{23 a b 1} \cdot 99+\overline{23 a}) \cdot 99+230+a+10 b+c+61= \\
& =(\overline{23 a b 1} \cdot 99+\overline{23 a}+2) \cdot 99+a+10 b+c+93
\end{aligned}
$$

thus 99 divides $a+10 b+c+93$. Since $a, b$ and $c$ are different digits that are not equal to 1, 2, 3, or 6, we have $a+10 b+c \geq 4+0+5=9$ and $a+10 b+c \leq 7+90+8=105$, so $102 \leq a+10 b+c+93 \leq 198$. Since $a+10 b+c+93$ is divisible by 99, it must be $a+10 b+c+93=198$, from which it follows that $b=9$ and $\{a, c\}=\{7,8\}$. Therefore, we have two solutions, $n=2379168$ and $n=2389167$.

![](https://cdn.mathpix.com/cropped/2024_06_07_6340f9febbabfa1859cdg-20.jpg?height=68&width=1585&top_left_y=2139&top_left_x=270)

Writing the number $n$ as $k \cdot 99+a+10 b+c+93$................................ 2 points Estimating $102 \leq a+10 b+c+93 \leq 198$ or using the estimate in writing the solutions .. 2 points Writing both solutions . .................................................................................. (If the contestant only writes both solutions, they get 1 point.)

2. method. Using the divisibility criteria for 9 and 11, we get that $2+3+a+b+$ $1+6+c=a+b+c+12$ must be divisible by 9 and $2-3+a-b+1-6+c=a-b+c-6$ must be divisible by 11. Since $a, b$ and $c$ are different digits that are not equal to 1, 2, 3, or 6, we have $a+b+c+12 \leq 9+8+7+12=36$ and $a+b+c+12 \geq 0+4+5+12=21$, so $a+b+c+12$ can only be 27 or 36. Similarly, $a-b+c-6 \leq 9-0+8-6=11$ and $a-b+c-6 \geq 0-9+4-6=-11$, so $a-b+c-6$ can only be $-11, 0$, or 11. From this, it follows that $(a+b+c+12)-(a-b+c-6)=2 b+18$ can only be 16, 25, 27, 36, 38, or 47. Since $2 b+18$ is an even number between 18 and 36, it must be 36, and from this it follows that $a+b+c+12=36$ and $a-b+c-6=0$. Therefore, $b=9$ and $a+c=15$. Since $a$ and $c$ are different from $b$, it must be $\{a, c\}=\{7,8\}$. Therefore, we have two solutions, $n=2379168$ and $n=2389167$.

Using the divisibility criterion for 9 and determining that $a+b+c+12$ is divisible by 9. 1 point Using the divisibility criterion for 11 and determining that $a-b+c-6$ is divisible by 11. 1 point

Determining that $a+b+c+12$ can only be 27 or 36. 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_6340f9febbabfa1859cdg-21.jpg?height=48&width=1579&top_left_y=844&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_6340f9febbabfa1859cdg-21.jpg?height=51&width=1579&top_left_y=891&top_left_x=273)

Writing both solutions ............................................................................................

(If the contestant only writes both solutions, they get 1 point.)