## Task $34 / 69$

We are looking for the smallest natural four-digit number $x$ (whose first digit is not zero) with the following property:

If in $x+1$ the two middle digits are swapped and the last (fourth) digit is subsequently removed, a three-digit number $y$ is formed such that $x$ is divisible by $11 y$ without a remainder.
Let $x=1000 a+100 b+10 c+d$ with $a \neq 0$. We distinguish two cases:

1. $d=0 ; 1 ; 2 ; \ldots ; 8$. Then $y=100 a+10 c+b$ and it follows the equation

$$
1000 a+100 b+10 c+d=11 k(100 a+10 b+c)
$$

and since the smallest $x$ is sought, $k=1$ must be; thus we get $100 b+d=100 a+100 c+11 b$. It is recognized that $d=b$ must be. The resulting equation $9 b=10 a+10 c$ is not solvable for non-negative $b$ and $c$ and positive $a$.

2. $d=9$. Then $y=100 a+10(c+1)+b$ and it follows the equation

$$
1000 a+100 b+10 c+d=11 k(100 a+10 c+10+b)
$$

also here we conclude $k=1$, thus $100 b+d=100 a+100 c+110+11 b$. In this case, $d=b$ must be, and it follows from $d=b=9$ the equation $7=a+c$.

Since the smallest number is sought, we infer $a=1, c=6$. The number $x$ sought is thus $\mathrm{x}=1960$ (the next larger number would be 2959).