Four pretty digits

Let the difference $d_1 - d_2 = d_3 - d_4 =: a$ .

A number is a multiple of 11 if and only if that is true for the alternating digit sum; that is, $d_1 - d_2 + d_3 - d_4$ must be a multiple of 11. This means that $(d_1 - d_2) + (d_3 - d_4) = 2a$ is a multiple of 11; but since for sure $|a| \leq 9$ this is only possible if $a = 0$ . However, that would make $d_1 = d_2$ and $d_3 = d_4$ , while we required that all digits are different.

The conclusion is that there are no such numbers.

Consider a 4-digit integer $N = \overline{d_1d_2d_3d_4}$ . Then

$\begin{aligned} N & \equiv 1000d_1 + 100d_2 + 10d_3 + d_4 \text{ (mod 11)} \\ & \equiv 10{\color{#3D99F6}d_1} + d_2 + 10{\color{#3D99F6}d_3} + d_4 \text{ (mod 11)} & \small \color{#3D99F6} \text{Let }d_1 - d_2 = d_3 - d_4 = n \\ & \equiv 10{\color{#3D99F6}(d_2+n)} + d_2 + 10{\color{#3D99F6}(d_4+n)} + d_4 \text{ (mod 11)} & \small \color{#3D99F6} \text{where } n \in [0, 9] \text{ is an integer.} \\ & \equiv 11(d_2+d_4) + 20n \text{ (mod 11)} \\ & \equiv 9n \text{ (mod 11)} \end{aligned}$

For $N$ to be divisible by 11, $n = 0$ or $d_1=d_2$ and $d_3=d_4$ . Therefore, there is no 4-digit integer of different digits that has the property. The answer is $\boxed 0$ .