I've noticed that if a number is divisble by 9, then the sum of the numbers is equal to 9. For example, 234 is divisble by 9. 2+3+4=9.
Another; 315 is divisble by 9. 3+1+5=9
What's the proof for this?
Ouch. How did I miss that!? However, if the digits add up to 9 it is divisible by 9. Like, 101001030003 is divisible by 9 because the digits add up to 9.
All I can conclude from my humiliating error is that it is not biconditional. You did it the other way round which proved false.
So it's conditional only. The converse is not true.
If the sum of the digits is a multiple of nine the the number is a multiple of nine.
EXAMPLE: $N=123456789$ has a digit-sum of $45$ therefore $N$ is divisible by nine.
P.S. The theorem is: $N$ is divisible by nine if and only if $N$ has a digit-sum divisible by nine.
Example: $99$ has a digit-sum of $18$ that is a multiple of nine.
$81276354$ is divisible by nine. WHY?
But repeating it, as Greg1313 said, works: 9+ 9= 18 and 1+ 8= 9.
This is called "casting out nines". If a number is NOT divisible by 9, say 57, then 5+ 7= 12 and 1+ 2= 3. 57 divided by 9 has remainder 3 87 8+ 7= 15 and 5+ 1= 6. 9 divides into 87 9 times with remainder 6.
This is sometimes used to check arithmetic calculations. Is 8392+ 7651= 16045? If we were to divide both sides by 9, 8392/9+ 7651/9. Is that equal to 16045/9? If the original addition is correct, then the sum of the remainders on the left should equal the remainder on the right. 8+ 3+ 9+ 2= 22, 2+ 2= 4. 7+ 6+ 5+ 1= 19, 1+ 9= 10, 1+ 0= 1. 4+ 1= 5. But 1+ 5+ 0+ 4+ 5= 15, 1+5= 6. No, that is not correct. Unfortunately, while it will tell you a calculation is wrong, it doesn't necessarily tell you when such a calculation is correct? Is 243*748= 190512? 2+ 4+ 3= 9 and 7+ 4+ 8= 19, 1+ 9= 10, 1+0= 1. 9*1= 9. 1+ 9+ 0+ 5+ 1+ 2= 18 and 1+ 8= 9. So it is correct? No, it is not. What is true is that 243*783= 190512. 748 and 784 have the same digits so the same digit sum but are, of course, different numbers.