I don't know how to explain this.

[DDBCAE CBAADCBE]

So I was messing around with letters and numbers making codes and I came upon something really weird that I'd like to get a second opinion on. I'm sure there has to be a mathematical explanation for this but for the life of me I just can't fathom it. I imagine it's something really simple that I'm just not thinking of. Specifically I'm talking about the 9s and 18s.

Spoiler: The thing in question

1 = 27 2+7=9
2 = 54 5+4=9
3 = 81 8+1=9
4 = 108 1+0+8=9
5 = 135 1+3+5=9
6 = 162 1+6+2=9
7 = 189 1+8+9=18
8 = 216 2+1+6=9
9 = 243 2+4+3=9
10=270 2+7+0=9
11=297 2+9+7=18
12=324 3+2+4=9
13=351 3+5+1=9
14=378 3+7+8=18
15=405 4+0+5=9
16=432 4+3+2=9
17=459 4+5+9=18
18=486 4+8+6=18
19=513 5+1+3=9
20=540 5+4+0=9
21=567 5+6+7=18
22=594 5+9+4=18
23=621 6+2+1=9
24=648 6+4+8=18
25=675 6+7+5=18
26=702 7+0+2=9
27=729 7+2+9=18
28=756 7+5+6=18

 

[Painted White]

If I understood what you're unsure about - It's quite simple actually. Any number with a sum of its digits is divisible by 3 if the sum is divisible by 3. This also applies to 9 if the sum is divisible by 9, but not any other number which is a multiple of 3 universally.
And, as 3 is the 3rd smallest number, there is a fuckton of numbers divisible by it.

100% of whole numbers are divisible by 1
50% divisible by 2
With 3 the % is not known to me nor easy to look up, but might just be 33 as far as I know, but it's incredibly high, and as such it's very easy to find a ton of numbers divisible by it (being a multiple of of it).

If we wrote out a list of all numbers divisible by 3, you'd find a ton of sums of 6, 9, 12, 15 etc.

That rule is there, but the proof and the way it is written and descibred is hard to understand for someone who's not good at math, and I sure as fuck am not so I'm not gonna link to it right now.

Source: Failed math in HS

 

[Clop]

You're surprised that the digital root of any number that's divisible by nine is a nine?

I don't understand what you're asking. And why are you posting any 18 at all when they're still a 9 in digital root? 1+8=9 by those rules, they're all nines.

 

[Not an NSA Plant 69]

You're surprised that the digital root of any number that's divisible by nine is a nine?

I don't understand what you're asking. And why are you posting any 18 at all when they're still a 9 in digital root? 1+8=9 by those rules, they're all nines.

Usually I see this in reverse and I have always referred to as the rule of 9s. It's most often used to determine if a number is a multiple of 9.

Simply stated any number that you can add the individual digits together and get 9 is a multiple of 9, if you get a sum with more than a single digit number as a result add those digits together and if you get a 9 it is a multiple of 9.

For example the sum of the digits 87,654,321 = 36, the sum of the digits 36 = 9 so 87,964,321 is a multiple of 9.

 

[Clop]

Usually I see this in reverse and I have always referred to as the rule of 9s. It's most often used to determine if a number is a multiple of 9.

Simply stated any number that you can add the individual digits together and get 9 is a multiple of 9, if you get a sum with more than a single digit number as a result add those digits together and if you get a 9 it is a multiple of 9.

For example the sum of the digits 87,654,321 = 36, the sum of the digits 36 = 9 so 87,964,321 is a multiple of 9.

87,654,321 and 87,964,321 are two different numbers. And the second is not divisible by nine, i.e. not a multiple of 9.

Yes, I know it's a typo, and I am shaming you for it.

 

[I should be working]

If you add numbers together for which the sum of those numbers are 9.
Then the sum of those numbers will be 9.
Mystery solved.

 

[Positron]

The proof is very simple actually, even more so if you know modular math.

There is a divisibility by 11 rule too. To prove it requires knowledge of modular math.

 

[qu_rahn]

how deep does this go