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

Reply Tue 26 Oct, 2010 02:21 pm
I was wondering if anyone has ever stumbled upon why the numbers 1-9 (and also 1-3) are whole divisible by 9 (and also 3) . As an example 123456789, 234567891, 2345678912, 56789123, 879143265, etc are all evenly divisible by 9.

123, 321,213,312,132 are all evenly disivible by 3.

Thanks - been one of those things oddities of the universe that has bothered me for years.
Bill Ellison
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Reply Tue 26 Oct, 2010 02:43 pm
@Bill Ellison,
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.
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Reply Tue 26 Oct, 2010 03:41 pm
@Bill Ellison,

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.
Reply Tue 26 Oct, 2010 08:14 pm
@Bill Ellison,
they be called factors
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Reply Wed 27 Oct, 2010 02:33 am
(...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").
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Reply Sun 12 Dec, 2010 01:39 am
@Bill Ellison,

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/

URL: http://able2know.org/reply/post-4393391

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