Any rectangular shape on a calculator numpad when divided by 11 gives an integer. Why? 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?

Start at the degenerate rectangle 1111, a multiple of 11.
Each time you move a side of your rectangle by one step in one of the 4 directions (leaving the other side in place), you add or remove one of these numbers (leading zeroes added for clarity):


*

*0011 (horizontally) or 0033 (vertically)

*0110 (horz.) or 0330 (vert.)

*1100 or 3300

*1001 or 3003


which are all multiples of 11 (obviously for the first three lines; 1001 = 11 * 91).
Adding such a number preserves being a multiple of 11.
This way from the degenerate rectangle 1111 you can reach any other rectangle (aligned to the numpad orientation), which prove they're all multiple of 11.
