A well known simplification.

If the sum of the digits is divisible by 3, then the number is divisible by 3. If the sum of the digits is divisible by 9, then the the number is divisible by 9.

So, if I gave you a number: 139527, then the sum of the digits is 1+3+9+5+2+7 = 27, which is divisible by both 3 and 9, so the original number is also divisible by 3 & 9. And, in the case above, the process can be repeated, so one can use the sum of the digits of the sum of the digits, or 2+7 = 9, divisible by both 3 and 9.

Add 3 to it, 139530, and the sum of the digits is 21 which is divisible by 3, but NOT 9.

I don't think that is true for 6, although, if a number is divisible by 3, and is an even number, then it is divisible by 6.

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Ok, mathematical proofs are a bit rusty. However, with a little hand waving, it is easy enough to demonstrate that this has to be true for factors of 9.

Consider you have an arbitrary number N that is divisible by 9.

Then N+9 is also divisible by 9.

But, N+9 = N+(10-1)

Hmmm???

Anyway, when adding 9 to any number from 9 to 81, one increases the 10's place by 1, and decreases the 1's place by 1, and the sum of the digits remains 9.

At 90+9, one is only incrementing it in the 1's place, but the rule would be essentially the same. Then you get: 99+9, 198+9, 297+9, ... 999+9?

Anyway, somehow there has to be a way to vigorously prove this for both the case of 3, as well as 9.