I have come across this fact a very long time ago, but I still can't tell why is this happening.

Given the standard calculators numpad:

```
7 8 9
4 5 6
1 2 3
```

if you dial any rectangular shape, going only in right angles and each shape consisting of 4 points, then the dialed number is always divisible by `11`

without a remainder.

Examples of the shapes: `1254`

, `3179`

, `2893`

, `8569`

, `2871`

, and so on. It is not allowed to use zero, only digits 1..9.

**UPDATE:** The part of the question below proved to be an error on my part because I did not double-check what the Programmers calculator was showing, and it turned out that it was rounding the results. *!!! See the accepted answer for an interesting follow-up to this, which actually expands the usecase to a working hexadecimal "keypad", and the other answers for even more various interesting approaches, discovering different aspects of this problem !!!*

~~The same rule also works even for the Programmer calculator layout on MS Windows 10 which looks like this:~~

```
A B 7 8 9
C D 4 5 6
E F 1 2 3
```

All valid rectangular shapes, for example, `A85C`

, `E25C`

, `39BF`

, and so on, being divided by `11`

still give an integer result!

Initially I was thinking that it's just somehow tied to picking digits from the triplets and being just another peculiarity of the decimal base and number `11`

and started looking this way, but discovering that it works for the hexidecimal base and even with the hex part of the keyboard layout not obeying exactly the pattern of the decimal part layout, I'm lost.

What law is this fun rule based on?