Any rectangular shape on a calculator numpad when divided by 11 gives an integer. Why?
Since you are concerned only with rectangular patterns, you have four digit numbers in a certain base $l$, and you want to check divisibility by $11$, where $11 = l+1$.
When you have a four digit number in base $l$, write it as $al^3 + bl^2 + cl + d$, where $0 \leq a,b,c,d < l$. (This represents the $l$-base number $\overline{abcd}$). Now, we have a cute fact : $l^3 + 1$ is a multiple of $l+1$, since $l^3+1 = (l^2 - l + 1)(l+1)$. Furthermore, $l^2 - 1 = (l-1)(l+1)$. Therefore, we make the following rewrite : $$ al^3 + bl^2 + cl + d = a(l^3 + 1) + b(l^2 - 1) + c(l+1) - (a - b + c - d) \\ = (l+1)(...) + ((b+d)-(a+c)) $$
where $l+1 = 11$ in base $l$.
Therefore, the remainder when $\overline{abcd}$ is divided by $11$ is $(b+d) - (a+c)$.
When you consider four numbers ($1 \to 9$, since I found problems in the letters for hex) in a rectangle and form a four digit number out of them now, can you see why this number $(b+d) - (a+c)$ is in fact zero, therefore giving your desired result?
Now that we have pointed out this rectangular pattern, I noted above that some counterexamples did exist for the MS - layout of hexadecimal numbers. The issue there was a fairly simple one : the "matrix" of entries did not satisfy the property that $a+d = b+c$ for $a,b,c,d$ going CW/CCW around any $2\times 2$ subrectangle of entries of the matrix.
If $l-1$ is not a prime, then we can actually arrange a "matrix" of $l$ entries which is not strictly column or row, but satisfies this "rectangle property" as we can call it. For this, write $l-1 = ab$ where $a,b \neq 1$, and arrange an $a \times b$ matrix of entries, which we fill in the following fashion : start with $1$ in the bottom corner, proceed towards right filling $2,3,...$ until you hit the end, then return to the left of the row above, and fill the next number, now repeat till you fill the whole matrix.
For $10$, this procedure yields the conventional keyboard pattern. For $15 = 5 \times 3$, it would yield $$ \begin{pmatrix} B&C &D & E& F\\ 6&7&8&9&A\\ 1&2&3&4&5 \\ \end{pmatrix} $$
which indeed will satisfy the property that any rectangle has $11$ in that base as a divisor. For example, $A8DF$, $1496$ and $CE97$ are all multiples of $11$.
Note that more is true : in fact, every parallellogram , read CW or CCW, leads to a multiple of eleven.
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.