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Multiples of 9

  1. Feb 23, 2005 #1
    Hi,
    I've a riddle that I can't solve. I hope someone can help me in it.

    Here is it:

    If we have a number and add the digits of that number, then subtract the summation we got from the original number, why is the result always multiples of the base system minus 1.

    Here is an example using the decimal system (base 10)

    The original number is 51, we add the digits, so we get 5+1=6, we subtract it from the original number we got 51-6=45 which is a multiple of 9 (Base 10 -1)

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    Another example in the octal system (base 8)

    We have 55, we add the digits we got 12 (octal) or 10 (decimal), we subtract so we have 55-12= 43 (octal) or 35 (decimal) which is a multiple of 7 (base 8 -1)

    I hope someone can help me in figuring out this puzzle.
    Thanks,
    Nadine
     
  2. jcsd
  3. Feb 23, 2005 #2

    arildno

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    Let us write a number N as follows:
    [tex]N=\sum_{i=0}^{M}a_{i}b^{i}[/tex]
    where b is the base.
    The sum S of the digits is evidently: [tex]S=\sum_{i=0}^{M}a_{i}[/tex]
    Hence, the difference D is:
    [tex]D=\sum_{i=0}^{M}a_{i}(b^{i}-1)[/tex]
    Does that clear it up for you?
     
    Last edited: Feb 23, 2005
  4. Feb 23, 2005 #3
    Thanks a lot :smile:
    I never thought it can be that simple.

    Thanks again,
    Nadine
     
  5. Feb 23, 2005 #4

    arildno

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    Welcome to PF, BTW!
     
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