Proof of divisibility(9)

1. May 10, 2012

Kartik.

Question - A proof of divisibility by the number 9 with a assumed number. The aim is to prove two assertions state within divisibility rule of 9 -
1. the sum of the digits of the number should be divisible by 9 and vice-versa

Attempt-
Let the number be N
N=a0a1a2a3..........an-1an (They do not represent the product but denotes the digits)
As in decimal number system -
N = an.10n+an-1.10n-1+......+a2.102+a1.n1
Where 10n = 9999.....9(n times)+1, and $n\geq1$
Then,
N= [an.99999....9(n times)+an-1.99....9(n-1 times)+.....+a2.99+a1.9]+(a0+a1+a2+a3+..........+an-1+an)

The first summand is definitely divisible by 9 but i am stuck over here to prove the remaining assertions, either verbally or i'm missing some computation too.

Last edited by a moderator: May 10, 2012
2. May 10, 2012

DonAntonio

What happened to $a_0\,$ here?

Is N a number divisible by 9? What direction are you trying to prove? You must be clearer.

You can try the following. Provide details:

=== Suppose N is a natural number divisible by 9. We must prove the final sum of its digits is 9.

So we can write $\,\,N=9k\,\,,\,\,k\in\mathbb{N}\,\,$ and induction of k: for k=1 this is clear, so assume for k and let us prove for k+1:

$9(k+1)=9k+9\,\,$ , and now you've the easy task to show that if the final sum of the digits of a numer is 9, then the same is true for the number + 9.

=== Suppose now N is a number s.t. that the final sum of its digits is 9. You must show N is a multiple of 9.

Now write $\,\,N=a_0+a_1\cdot 10+...+a_n\cdot 10^n\,\,so\,\,F(a_0+a_1+...+a_n)=9\,\,$ , with the F indicating the final digit sum, and now write $$N=a_0+a_1(9+1)+a_2(99+1)+...+a_n((10^n-1) +1)$$ Continue from here, proving first that for any $\,\,n\in\mathbb{N}\cup\{0\}\,\,,\,\,10^n-1\,\,$ is a multiple of 9.

DonAntonio

Last edited by a moderator: May 10, 2012