B4. $\overline{a b}$ is the notation for the number written with the digits $a$ and $b$, where $a \neq 0$.

Give all positive integer values of $K$ for which the following holds:

- $K$ is a positive integer
- there exists a number $\overline{a b}$ that is not divisible by 9 with $\overline{a b}=K \times(a+b)$.

Note: Points will be deducted for incorrect values of $K$!
B4 The second condition can be rewritten as $10 \times a + b = K a + K b$ which simplifies to

$(10 - K) \times a = (K - 1) \times b$. Since $K$ must be a positive integer, we only need to examine the values from 1 to 10 for $K$. Additionally, it must hold that $1 \leq a \leq 9$ and $0 \leq b \leq 9$.

If $K=1$ then $a=0$, which is not allowed. Therefore, $K=1$ does not satisfy the condition.

If $K=2$ then $8 \times a = b$ and thus $a=1, b=8$ and $\overline{a b}=18$, which is divisible by 9. Therefore, 2 does not satisfy the condition.

If $K=3$ then $7 \times a = 2 \times b$ so $a=2, b=7$ and $\overline{a b}=27$, which is divisible by 9. Therefore, 3 does not satisfy the condition.

If $K=4$ then $6 \times a = 3 \times b$ so for example $a=1, b=2$ and $\overline{a b}=12$, or $a=2, b=4$ and $\overline{a b}=24$. 12 and 24 are not divisible by 9. (For example, $12=4 \times(1+2)$.) Therefore, $K=4$ is a possible value.

If $K=5$ then similarly $\overline{a b}=45$ and that is divisible by 9. Therefore, 5 does not satisfy the condition.

If $K=6$ then similarly $\overline{a b}=54$ and that is also divisible by 9. Therefore, 6 does not satisfy the condition.

If $K=7$ then $3 \times a = 6 \times b$ so for example $a=2, b=1$ and $\overline{a b}=21$, or $a=4, b=2$ and $\overline{a b}=42$. 21 and 42 are not divisible by 9. (For example, $42=7 \times(4+2)$.) Therefore, $K=7$ is a possible value.

If $K=8$ then only $\overline{a b}=72$ and that is divisible by 9. Therefore, 8 does not satisfy the condition.

If $K=9$ then only $\overline{a b}=81$ and that is also divisible by 9, so 9 does not satisfy the condition either.

If $K=10$ then $b=0$. This gives solutions, for example, $30=10 \times(3+0)$. Therefore, 10 is another value that satisfies the condition.

In summary: We find 4, 7, and 10 as possible values for $K$.

(C) Stichting Nederlandse Wiskunde Olympiade 2006

![](https://cdn.mathpix.com/cropped/2024_04_17_a2053d8a9f115fa3d794g-2.jpg?height=389&width=415&top_left_y=2107&top_left_x=1343)