4. Let $A$ and $B$ be $n$-digit numbers, where $n$ is odd, which give the same remainder $r \neq 0$ when divided by $k$. Find at least one number $k$, which does not depend on $n$, such that the number $C$, obtained by appending the digits of $A$ and $B$, is divisible by $k$.
Solution. Let $A=k a+r, B=k b+r, 0<r<k$; then we have:

$$
C=10^{n} A+B=10^{n}(k a+r)+(k b+r)=k\left(10^{n} a+b\right)+\left(10^{n}+1\right) r .
$$

The number $C$ is divisible by $k$ if and only if the number $10^{n}+1$ is divisible by $k$. Since $n=2 m+1$, we have

$$
10^{n}+1=10^{2 m+1}+1=(10+1)\left(10^{2 m}-10^{m-1}+\ldots+1\right)=11 D
$$

which means that a number $k$ that satisfies the condition of the problem is the number 11.

## 4th year