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A Puzzle about a Calculator

31 pointsby surprisetalk5 months ago

7 comments

qrian5 months ago
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
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gorkish5 months ago
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.<p>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.
teucris5 months ago
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?
russdill5 months ago
Weird, every time I try, I just get &quot;A suffusion of yellow&quot;
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lilyball5 months ago
Interestingly a 45º rotated rectangle using the keys 4 8 6 2 also is divisible by 11. This isn&#x27;t directly addressed in the solution, although if you change &quot;move both numbers horizontally or vertically by the same distance&quot; to say &quot;and&quot; instead of &quot;or&quot; then it does.
jahbrewski5 months ago
How do you &quot;read&quot; an article like this? I would need to pull out some paper, run calculations, etc. to understand this (but perhaps I&#x27;m not the intended audience, as a non-mathematician?) - Or is that how you all approach an article like this?
d_tr5 months ago
&gt; If you allow zero width&#x2F;height, you can also have 7777, 7887, 7997, 7447 and 7117.<p>Why not 7227, 7337, 7557, 7667 too?<p>The theorem holds for these as well.
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