Algebra - Numbers Divisible by 9 (Friendly Version)

How do you know when a number is divisible by 9? If the sum of its digits is divisible by 9, then the original number is divisible by 9! For example, 6525 is divisible by 9 because the sum of its digits (6 + 5 + 2 + 5 = 18) is divisible by 9 (18/9 = 2). But, how and why does this happen?9 times 1 is 9, and 9 is divisible by 9; 9 times 2 is 18, and 18 is divisible by 9; 9 times 3 is 27 and 27 is divisible by 9; so, in general, 9 times any number is divisible by 9. 9x is divisible by 9; 9y is divisible by 9. It doesn't matter; x can be a number or a long sum of numbers; just as long as you multiply by 9, you can always divide a 9 out. Let us remember 9x. We want to get a number in the form of 9x. Then we know it has to be divisible by 9.Take 6525 for example! Now ask yourself if the mathematical sentences are true or not.

Is sentence below true?

6525 = 6(1000) + 5(100) + 2(10) + 5(1)

Yes; look at the metric system: 6m + 5dm + 2cm + 5mm = 6525mm

Is sentence below true?

6525 = 6(999 + 1) + 5(99 + 1) + 2(9 + 1) + 5

Again, yes because 10 = 9 + 1; 100 = 99 + 1, and so on.

Is the sentence below true?
6525 = 6(999) + 5(99) + 2(9) + 6 + 5 + 2 + 5

Yes, because we are allowed to distribute numbers like this. For example, x(x + 2) = x^2 + 2x, or 25 = 5(5) = 5(2 + 3) = 5(2) + 5(3) = 10 + 15 = 25. Now notice that the numbers on the right are the sum of the number 6525, which is 6 + 5 + 2 + 5 = 18, and 18 is divisible by 9 because 9 times 2 =18, and again 9 times any number is divisible by 9.

Is this sentence below true?
6525 = 9(111)(6) + 9(11)(5) + 9(2) + 9(2)

Yes, because all we did was factor out a 9 from each term. The last term was the sum of the digits 6525.This sentence below is true because we can factor a 9 from each term, we couldnt have done this if the sum of the digits in 6525 was NOT divisible by 9, so we need the sum of the digits to equal a number that's divisible by 9 in order to factor it out at this step!

6525 = 9[6(111) + 5(11) + 2(1) + 2]

Now remember, we said 9 times ANY NUMBER is divisible by 9, well let's make that number [6(111) + 5(11) + 2(1) + 2], now we have shown not only how but why if the sum of digits are equal to a number divisible by 9, that number MUST be divisible by 9, and vis versa. Here are all the steps in order:

6525 = 6(1000) + 5(100) + 2(10) + 5(1)6525 = 6(999 + 1) + 5(99 + 1) + 2(9 + 1) + 5
6525 = 6(999) + 5(99) + 2(9) + 6 + 5 + 2 + 5
6525 = 9(111)(6) + 9(11)(5) + 9(2) + 9(2)
6525 = 9[6(111) + 5(11) + 2(1) + 2]

6525/9 = 6(111) + 5(11) + 2(1) + 2725 = 666 + 55 + 2 + 2 = 725

Abstract Algebra - Numbers Divisible by 9 (Technical Proof)

This is a technical proof of how if you sum the didgets of any number and that sum is divisble by 9, then the original number is divisible by 9.

Let ai denote a digit in mod 10 and let a1a2…an denote a positive integer with an n number of digits. Then, 9a1a2…an Î Z+ if and only if 9Si = 1, n ai Î Z+.

Proof. Factoring out a 10n from the nth digit gives

a1a2…an – 1an
= 10(n – 1)a1 + 10(n – 2)a2 + … + 10(n – (n – 1))an – 1 + 10(n – n)an
= 10(n – 1)a1 + 10(n – 2) + … + 10an – 1 + an

Since [(10n – 1) + 1] = 10n,

= [(10n – 1 – 1) + 1]a1 + [(10n – 2 – 1) + 1]a2 + … + [(10 – 1) + 1]an – 1 + an.

Distributing ai to each term gives

= (10n – 1 – 1)a1 + a1 + (10n – 1 – 1)a2 + a2 + … + (10 – 1)an – 1 + an – 1 + an

Because 9(10n – 1) "n Î Z+, 9 can be factored out from each (10n – 1) to leave (S i = 0, n 10i).

= 9a1(S i = 0, n – 1 10i) + 9a2(S i = 0, n – 2 10i) + … + 9an – 1 + a1 + a2 + … + an (5)

The number a1a2…an – 1an contains the sum Si = 1, n ai. Let 9Si = 1, n ai = R. Factoring out 9 from each term gives

a1a2…an – 1an = 9a1(S i = 0, n – 1 10i) + 9a2(S i = 0, n – 2 10i) + … + 9an – 1 + 9R
= 9[a1(S i = 0, n – 1 10i) + a2(S i = 0, n – 2 10i) + … + an – 1 + R]

Because a1a2…an – 1an can factor out 9, 9a1a2…an – 1an. This could not have occurred if R did not factor out nine as well. Further, if 9a1a2…an – 1an, then a1a2…an – 1an gives a 9R term by equation (6). 9R = Si = 1, n ai and 9Si = 1, n ai. QED