Why, when I divide any number, and a scrambled variation of that number, by 9 do the two results have the same remainder?

Answer 1

You can rewrite any number in our numerical system this way:

a_na_n-1a_n-2....a₀= 10ⁿa_n+10^n-1a_n-1+10^n-2a_n-2 + ...+ 10a₁ + a₀=

=(9999....9*a_n+ 999....9*a_n-1+99....9*a_n-2 + ...+9a₁ ) + (a_n+a_n-1+a_n-2 + ...+ a₁ + a₀)

So, basically we can separate any number in two sums, the first one is a multiple of nine and the second one is the sum of the digits of the numbers. Then it doesn't matter the order of the digits in the number, you'll obtain the same answer for the second sum, therefore when you divide by 9 you'll have the same remainder.

To show this with a numerical example:

571344= (99999*5+9999*7+999*1+99*3+9*4) + (5+7+1+3+4+4)

dividing both sides by 9 you get:

571344 / 9 = (11111*5 + 1111*7 +111*1 + 11*3+ 1*4) + (5+7+1+3+4+4) / 9

Conclusion: The remainder of the division by 9 depends only on the sum of the digits of the numbers, and the order doesn't matter.