Calculators

[patdunham]

On a standard calulator keypad, if you take any block of 4 keys and key then in, either clockwise or anticlockwise, starting with any key, the answer is always divisible by 11. E.g. 4785 or 2563. This is also true for the 8 circumference keys. Does anyone know of a mathametical reason why this should be?

[dundurn]

The diagonally opposite corners of a "rectangle" of numbers always add up to the same total as the other pair of diagonally opposite numbers - e.g. 1+5 = 2+4. When you write these numbers down as a 4 digit number whether clockwise or anticlockwise then the resulting number always has the following property. Subtracting the sum of the first and third numbers from the sum of the second and fourth numbers equals 0. Such a number is always divisible by 11.

[patdunham]

Thanks for that

[sidkid]

Well .....I understand your explanation right up until the very last line - why is such a number always divisible by 11?

[patdunham]

If you reverse the question, any number abc when multiplied by 11 becomes the sum of that number x10 plus itself so: abc0 + abc
abc0
+ abc
--------

so the sum of the first and third digits of the answer will be a+b+c and the sum of the 2nd & 4th digits will be a+b+c+0, so any 4 digit number meeting the criteria that 1st + 3rd = 2nd + 4th must be divisible by 11

[dundurn]

One thing to add. For the purpose of the question the definition of a number divisible by 11 (sum of alternate digits = sum of other alternate digits) was sufficient. Numbers where the difference betweeen the two answers = 11 also are divisible by 11 e.g. 10901

[patdunham]

thanks to everyone