Cfrgtkky

Here is a more elaborate description of the problem statement. What I found with a few examples is that given a number, say 569. If we are required to sum its digits repetitively until the sum is single digit, i.e

1)5+6+9 = 20;

2)2+0 = 2

where 2 is single digit sum

We can compute the result, 2, with just one run through the number as follows:

1) 5+6 = 11 // adding first 2 digits of 569

2) 1+1 = 2 // adding the sum of 5,6 since its more than 1 digit

3) 2 + 9 = 11 // adding the sum of digits of 11 to 9, the last digit of 569

4) 1 + 1 = 2 // summing the digits of result of step 3

which yields the same result, 2.

Could someone prove mathematically(or otherwise) why this works?
OR
provide an example where it won't work? I'm pretty sure it works all the time but don't have a proof..

Edit 1:
@Mark Bennet. Thanks for your answer. I was able to fill in the gaps and understand why the answer can be derived directly from the initial positive number using modulo 9. I will write out the proof after telling what I didn't understand.

Let original number be n.
I understood that the answer would be x = n%9 if x > 0 and 9 if x = 0;

However, what I didn't understand is, why when going by the less efficient way of repetitively summing the digits of the numbers, the order of adding the digits does not matter.

Proof of what I got from your explanation:

Let n be original number, s1, s2, s3, ..., sk be the sums formed by adding the digits. Let sk be single digit.
We know what n-s1, s1-s2, s2-s3 are all divisible by 9 because of your explanation. So we can add them all up, cancelling s1,s2, ..., sk-1j, until we are left with just n-sk which is also divisible by 9.

So, we can write n = 9p+sk where p >=0. By definition, sk is n%9.

What I didn't understand, (or rather connect this proof to) what my question is initially seeking an answer for: If I still went by my way of adding digits repetitively, why does the order of adding digits not matter?

Think about just one digit. In the original number it appears as $atimes10^r$ because of its "place value". In the sum of digits it appears just as $a$.

The difference between the two is $atimes (10^r-1)$, and you will see that $10^r-1$ has every digit $9$ (or is zero if $r=0$) and is therefore divisible by $9$.

The difference between the original number and its sum of digits is therefore divisible by $9$, and this can be used to show that when you start with a positive integer the single digit you get at the end is $9$ if the number is divisible by $9$ or is otherwise the remainder you get when you divide the original number by $9$.

It doesn't matter how this sum is calculated - beginning with a single digit shows that this can be done digit by digit. The difference at every stage between the number you started with and the one you finish with is divisible by $9$. That is the key point.

I have left some gaps for you to fill in. There are plenty of slick proofs out there ("casting out nines" is an old name for this, which was a check on arithmetic in the days before electronic calculators), but you will do best to work it through for yourself.