Any parellogram would work since given four digit number n1n2n3n4, it is divisible by 11 iff n1+n3=n2+n4, and each ni is linear combination of the coords of keypads xi, yi, and thus (n1+n3)/2 = (n2+n4)/2

Nice - the n1 + n3 = n2 + n4 equality is only necessary (mod 11) e.g. 9020 works - this is because 99...99 with even # of 9s is divisible by 11 and with odd # 9s is divisible by 11 if we subtract 9 (or add 2) so then is = -2 mod 11. So then for example with 4 digits

  1000a + 100b + 10c + d = [a + b + c + d] + [999a + 99b + 9c]
                         = [a + b + c + d] - 2a - 2c (mod 11)
                         = (b + d) - (a + c) (mod 11)

For a 4 by 4 hexadecimal keypad with 0x0..0xF keys in any sequential row or column arrangement the same will be true that all possible numbers will be a multiple of 0x11.

This holds true for any size square keypad; the common multiple will be 1+x^2 and for square-number bases, the value of the common factor will always be “11” in conventional symbology. A 2x2 base 4 and even a 1x1 binary keypad respects the rule although it’s sort of meaningless in the latter context.

At first I thought this was going to be a puzzle about getting to a specific number using certain rules for navigating the pad, including the operation buttons. For instance, by pressing one or two buttons in each row from top to bottom, can you get the calculator to display 70?

Weird, every time I try, I just get "A suffusion of yellow"

That's a known bug whenever the calculated result is > 4.0. Just scale down your inputs/outputs and it will work fine :)

(for those who don't know, the original comment is a reference to the novel "The Long Dark Tea-time of the Soul" by Douglas Adams)

Interestingly a 45º rotated rectangle using the keys 4 8 6 2 also is divisible by 11. This isn't directly addressed in the solution, although if you change "move both numbers horizontally or vertically by the same distance" to say "and" instead of "or" then it does.

How do you "read" an article like this? I would need to pull out some paper, run calculations, etc. to understand this (but perhaps I'm not the intended audience, as a non-mathematician?) - Or is that how you all approach an article like this?

> If you allow zero width/height, you can also have 7777, 7887, 7997, 7447 and 7117.

Why not 7227, 7337, 7557, 7667 too?

The theorem holds for these as well.

I assume it has to do with them being rectangular. And while technically you could have a rectangle that is rotated, I think it's also plausible to assume that any diagonals are considered as rhombuses (rhombi?)

I would call 2486 and 2684 a square and, hence, a rectangle. I think the theorem holds for all 8 variants of that.

2486 is only a square/rectangle if the keys are exactly square. They're certainly not in the example image shown.

Also 7007.