Why are the numerals 1-9 always wholely divisible by 9?

First, ask yourself why numbers where the digits add up to a multiple of nine are divisible by nine. Let's take the example

a+b+c = multiple of nine, then is 100a+10b+c is divible by nine?
100a + 10b + c = (99a + 9b) + (a + b + c) = 9 (11a + b) + (a + b + c)

The first term is clearly divisible by nine, so if (a+b+c) is divisible by nine, then the entire number is. The same technique works for numbers divisible by three. You can see that this will work for any number of digits. For example, four digits would yield

9(111a + 11b + c) + (a + b + c + d)

So back to your original question, any collection of digits that adds up to a multiple of nine can be arranged in any manner and will still be divisible by nine. It just so happens that the sum of the digits 1-9 is 45, so it always works.

Alternatively

In the number abcde...etc, all letters to the left of e represent bunches of 10's, and all such bunches will have a remainder of 1's equal to its letter when divided by nine. Thus all the left over 1's will be the total of all the letters, including e. So If the total of those remainders divides by nine, the whole number does. (Ditto 3's)

And as engineer said the sum of the digits 1-9 divides by 9, irrespective of the order you write abc...etc.

they be called factors

(...just to clarify "bunch of 1o's method", take as an example 7000 which equals 700 tens. Divide that by 9 and you get 700 ones as first remainder answer, but divide that by 9 and you get 70 ones, leading to 7 ones as the final remainder for that "bunch").

Hello Bill,
Since I've met this question only know, I do not know if it is actual for you. Maybe you've got an answer.
However, I use thje theorem: If the sum of the digits of the number can be divided by 9 or 3 without a reminder, the numver is all divisible by 9 (and also 3) . I can supply you a proof if you want.
In our case, the digits' sum of each of those types of number is: 1+2+3+4+5+6+7+8+9=45 divisible by 9 (and also by 3)
Therefore each of those numbers can be divided by 3/9 If you've questions, you can send ,me a mail Amos [email protected] URL: http://able2know.org/reply/post-4393391 http://able2know.org/reply/post-4393391 http://able2know.org/forum/math/

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