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well r/math took it down so heres my shot

6 Answers

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This is well known. Any number which we would write as “ABC”, where A, B and C are the digits (between 0 and 9), is actually 100 * A + 10 * B +C . We can rearrange this as:

 (99 * A +9 * B) + (A+B+C)

Now the first bracket is always divisible by 9. So if the whole number is divisible by 9, A+B+C must be as well. And the reverse is also true. You can extend this to any number of digits.
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It's a fairly well known and relatively easy-to-prove rule, and more generally can be stated as:

> The sum of the digits of a base-10 representation of a number n is congruent to n (mod 9).

This has the added effect of being true mod 3, because of course if you have a multiple of 9, or 3 more than a multiple of 9, or 6 more than a multiple of 9, you necessarily have a multiple of 3.

Just google "divisibility by 9" and there will be a bunch of discussions and proofs that show up. I'm not sure if it has any formal name beyond that.

You may be interested in other such divisibility rules, which often get brought up in discussions about this particular rule. Feel free to look up divisibility rules for things like 11 as well.
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It's not clear what question you're asking.
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thanks to every one for your comments on this i appreciate it a lot but what i said also works with larger numbers such a 1+1+1+1+1+1+1+1+1= 9 therefore the number 111111111 is divisible by 9 and say you number is 222222222 (2+2...) is = to 18 and do the thing again 1+8 meaning 222222222 is divisible by 9 so any number that can have its seprate digits add up to 9 or a multiple of 9 is there for a multiple of 9
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Interestingly, I enjoy playing with numbers a lot :)

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