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

  1. #1
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    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 9 then the number is divisible by 9. E.g. the number 72 the sum is equal to nine ( 7 + 2 = 9) and we know for a fact that 8 \cdot 9 = 72.

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

    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 9, we can write

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

    But now I'm stuck. The proof is complete if we can somehow factor the number N by breaking out a 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 10 = 9 +1.
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  2. #2
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    Re: Proving the sum of digits is always 9 if number is divisible by 9

    I am gonna try modular arithmetic...
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  3. #3
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    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!
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    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.
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  5. #5
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    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.
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