Results 1 to 5 of 5

Thread: Proving the sum of digits is always 9 if number is divisible by 9

  1. #1
    Junior Member
    Joined
    Jul 2012
    From
    Earth
    Posts
    56

    Proving the sum of digits is always 9 if number is divisible by 9

    The other day, I was attempting to prove that if the sum of all the digits equal to $\displaystyle 9$ then the number is divisible by $\displaystyle 9$. E.g. the number $\displaystyle 72 $ the sum is equal to nine ($\displaystyle 7 + 2 = 9$) and we know for a fact that $\displaystyle 8 \cdot 9 = 72$.

    Let $\displaystyle N$ be an arbitrary number with an arbitrary number of digits. In expanded form, we can write

    $\displaystyle N = a \cdot 10^b + c \cdot 10^d + e \cdot 10^f + \, \ldots \, + y \cdot 10^1 + z \cdot 10^0 \, .$

    Now, assuming that the sum of all the digits equals to $\displaystyle 9$, we can write

    $\displaystyle 9 = a + c + e + \, \ldots \, + y + z$

    But now I'm stuck. The proof is complete if we can somehow factor the number $\displaystyle N$ by breaking out a $\displaystyle 9$ but I don't really see how I'm supposed to go about to achieve that. The binomial theorem doesn't prove very fruitful (I think) when substituting $\displaystyle 10 = 9 +1$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jul 2012
    From
    Earth
    Posts
    56

    Re: Proving the sum of digits is always 9 if number is divisible by 9

    I am gonna try modular arithmetic...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2012
    From
    Earth
    Posts
    56

    Re: Proving the sum of digits is always 9 if number is divisible by 9

    Yep, modular arithmetic was the key, I managed to prove it!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,767
    Thanks
    3027

    Re: Proving the sum of digits is always 9 if number is divisible by 9

    Though it is really the same as working "mod 10" you can show directly that 10^n divided by 9 has remainder 1 for any n. 10^n= 999...9 (with n "9"s) plus 1.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2012
    From
    Jaipur,Rajasthan
    Posts
    36
    Thanks
    1

    Re: Proving the sum of digits is always 9 if number is divisible by 9

    Sorry bro.If you solve it then let me know.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Feb 20th 2013, 09:32 AM
  2. Replies: 4
    Last Post: Jul 18th 2011, 06:16 PM
  3. Replies: 2
    Last Post: May 3rd 2011, 03:22 PM
  4. Replies: 7
    Last Post: Nov 28th 2010, 09:22 PM
  5. number divisible
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Aug 5th 2008, 07:25 PM

Search Tags


/mathhelpforum @mathhelpforum