The magnificent number nine

Written by Kjetil Olseng, Anneli Nesteng and Øystein Eriksen

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Once upon a time

man discovered number nine.

If you play the telephon game,

you`ll always end up with the same;

If you add the digits, you`ll get nine,

no matter how you putèm in a line

If you think of a phonenumber, for example 354276, and mix up the figures (573642), then subtract the smallest nuber from the larger one, and then you add the figures together, you`ll get 27. If you add 2+7 you`ll get nine.

This is because the remainder nine merely dependes on the sum of the figures; therefore, it does not change if we mix up the digits. If we subtract two numbers whose remainder is the same, then the result will be divisible by nine, and, therefore, the sum of the digits can also be divided by nine.

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We can begin with an example: If you the 53, so can say that if this is 10a+b, then a is 5, and b is 3. You should subtract the same number turned around, in our case 35. To get this with a and b, you must say that 35 is 10b+a. You can set up like this: (10a+b) - (10b+a)=

10a + b - 10b - a =

9a-9b=

9 (a + b)

Here you get nine as a factor, and therefore will your answer always something with 9

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If you have a three digit number, for example 634, and then you subtract its reversal, 436, from that number, the answer you’ll get will always have nine as a common divisor, and the sum of the digits will be nine. In our example:

634

-436

=198

198 has nine as a divisor, and the sum of the digits will at last be nine (1+9+8=18; 1+8=9).

Explanation:

You can call the digits a, b and c (abc=634; a=6, b=3, c=4). Note: to make this trick work: a>c

The sum of the digits in abc is then 100*a + 10*b + c, and the reversal, cba, is 100*c + 10*b + a:

(100a + 10b + c) - (100c + 10b + a)=

100a + 10b + c - 100c - 10b - a =

99a - 99c = 99 (a - c)

99 (and then 9) will always be a common divisor, and this also means that the sum of the digits will be nine.