[CZARCAR (OP)]

consider the multiples of 3 and there are tables online. all multiples of 3 can be resolved to 3, 6, or 9, buy simply adding the factors of the multiples to result in 1 number. consider the multiple of 3 which is 666... 6 + 6 + 6 equals 18... and 1 + 8 equals 9. this silly message whatever it is called seems to resolve all multiples of 3 to either 3, 6, or 9. Why? I call it astrological arithmetic because it popped into my head when I was drunk AA

[CliffordK]

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.

============

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.

[Lmnre]

This phenomenon involves the process of casting out nines.

Casting out nines works because, in our base 10 system, 9 = 10 – 1, and when the digits in a number are treated as the same (and thus are additive), then the above equation becomes, 9 = 1 – 1 = 0, and so, casting out nines does not effect the summation of the digits.

When we look at the multiples of 3, we find 3, 6, 9, 12, 15, 18, ... . We see the 3, 6, and 9 in the first three multiples, and we start casting out nines for 12, 15, and 18, which gives us 3, 6, and 9. Or, to use the shortcut method you mentioned, 1+2=3, 1+5=6, and 1+8=9.

[Bored chemist]

I asked my 12 fingered friend from the planet Zog and he says this doesn't work for him.

[CliffordK]

I asked my 12 fingered friend from the planet Zog and he says this doesn't work for him.

Yet, it does work with a base 12 system, just multiples of 11, which happens to be prime.


 0  1  2  3  4  5  6  7  8  9  A  B
10 11 12 13 14 15 16 17 18 19 1A 1B
20 21 22 23 24 25 26 27 28 29 2A 2B

If you did it with B (11).

1xB = B
2xB = 1A 1+A = B
3xB = 29 2+9 = B

BxB =  A1   A+1 = B
Bx10 = B0   B+0 = B
Bx11 = BB   B+B = 2B

[CZARCAR (OP)]

thank you