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Thread: Sums of digits in multiples

  1. #1 Sums of digits in multiples 
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    In positive multiples of 9 (e.g. 18, 72, 909) why do the sum of the digits (no matter what order or combination) make 9 or another multiple of 9?

    9 = 9 = 9
    18 -> 1+8 = 9
    27 -> 2+7 = 9
    909 -> 9+0+9 = 18 (1+8=9) or... 90+9=99 (9+9=18, 1+8=9)

    So no matter how large the multiple is, you can deduce it by summing its digits until you eventually reach 9.

    Also, is there a way to prove this for every multiple of 9?


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  3. #2  
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    Its true. The gneral idea of a proof is that every power of ten is 1 + multiple of 9.
    For example a 3 digit number abc is (99 + 1)a + (9 +1)b +c. Since 99a + 9b is obviously divisible by 9, then a+b+c must be divisible by 9.


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    Okay, I see now. Thank you. I was at first confused with your statement, considering 28 (9*3+1) is not a power of ten. But it must have had that "all m's are n's, but not all n's are m's" quality.
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    Alternatively, it's called "modulo 9 arithmetic", which is just another way of saying "base 10 arithmetic". If you want the full works, the modulo n relation partitions the integers into disjoint modulo classes, for which each member of one class differs from some member of another class by n or a multiple thereof.

    Try this party trick:

    Invite your victim to chose a number with more than one digit, add them (and add again if needed) to obtain a single digit.

    Now say to subtract this number from the starting number, and repeat the adding digits process until he arrives again at a single digit.

    Close your eyes, pretend to think really hard and state "the answer is 9".

    For any cheats who start with a single digit, the answer is of course zero.
    Last edited by Guitarist; December 21st, 2011 at 12:22 PM.
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  6. #5  
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    Quote Originally Posted by brody View Post
    In positive multiples of 9 (e.g. 18, 72, 909) why do the sum of the digits (no matter what order or combination) make 9 or another multiple of 9?

    9 = 9 = 9
    18 -> 1+8 = 9
    27 -> 2+7 = 9
    909 -> 9+0+9 = 18 (1+8=9) or... 90+9=99 (9+9=18, 1+8=9)

    So no matter how large the multiple is, you can deduce it by summing its digits until you eventually reach 9.

    Also, is there a way to prove this for every multiple of 9?
    Consider the number where each of the is a digit from 0 to 9.

    Then where .

    Notice that for each non-negative integer , is divisible by 9:


    This can be rigorously established by mathematical induction (or by simply noting that ).

    It follows that is divisible by 9. Hence we have , i.e. is divisible by 9 if and only if the sum of its digits is divisible by 9.
    Last edited by Nehushtan; March 4th, 2013 at 08:48 AM.
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  7. #6  
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    the 9s repeat in the order of 1/9 until inf
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  8. #7  
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    Quote Originally Posted by brody View Post
    So no matter how large the multiple is, you can deduce it by summing its digits until you eventually reach 9.

    Also, is there a way to prove this for every multiple of 9?
    You could also look at a generator function. and write the rules about the digits

    Multiples of 9 are just 9+9+9 for n 9's.
    So use addition.

    When will adding 9 break the rule that the sum of digits is a multiple of 9, and recursive summing of digits will not end at 9 ?


    If the right hand digit is 0, then you change the 0 to 9, which does add 9 to the sum of the digits.
    Is this leaving the sum of digits as a multiple of 9 ? yes.


    Every time you add 9, you take 1 off the right hand digit, and add 1 to the digit to the left.
    If you have not caused a carry, then there is no change to the sum of the digits, since you took 1 off one digit and added 1 to another.
    So this is leaving the sum of the digits to be a multiple of 9.

    If you have caused a carry, this is because 9+1=10, which means you remove a 9 from one column, and and taken 1 off one column and added 1 to another.. The total change in the sum of digits, but the sum of digits must remain a multiple of 9.

    Even if there are multiple 9's eg "xxx999992" , the carry will eventually come to a column where the 1 can you still do a carry of the to the next digit left.

    Now since you increased through 2 digits, 3 digits, etc, its sure that the smaller ones were always meeting the rules, because they were generated according to the rules.

    Also, since its a number larger than 0, the digits cannot sum to exactly 0. The removal of the 9 could never reduce the sum of digits to be 0.


    Therefore, its not possible that the digits of a multiple of 9 do not sum to a multiple of 9.
    And recursive summing of the digits must always produce lesser and lesser sums, which much be multiples of 9, and so when it reaches a single digit, that digit must be 9
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    I agree if an only if it is a whole number multiple of 9. Given a fractional multiple of 9 say z/9 multiplied by 9 then an identity is created and any given number z is returned even ones that dont add to nine. Even then I havent tested every whole number multiple of nine greater that zero to prove that a whole number multiple which doesnt add to nine doesnt exsist.
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  10. #9  
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    Quote Originally Posted by brody View Post
    In positive multiples of 9 (e.g. 18, 72, 909) why do the sum of the digits (no matter what order or combination) make 9 or another multiple of 9?

    9 = 9 = 9
    18 -> 1+8 = 9
    27 -> 2+7 = 9
    909 -> 9+0+9 = 18 (1+8=9) or... 90+9=99 (9+9=18, 1+8=9)

    So no matter how large the multiple is, you can deduce it by summing its digits until you eventually reach 9.

    Also, is there a way to prove this for every multiple of 9?

    All the multiples of 9 will end up reaching to 9 if the result is added as per the example given.

    108 = 1+0+8=9
    117=1+1+7=9 and so on.

    the rule applies for all the multiples of 9.
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