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Number minus sum of digits is dividable by 9

  1. Aug 11, 2011 #1
    Hello!
    Why is it so, that if you have a number, for example 183, you take the sum of the digits of that number, so 183-(1+8+3), u get a number that is always dividable by 9?

    I'm sorry if this is a dumb question, I'm not very good at math but still curious to know.
     
    Last edited: Aug 11, 2011
  2. jcsd
  3. Aug 11, 2011 #2

    phinds

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    You might find it interesting to look into the concept of positional notation. This kind of problem is trivial if you think about the underlying number system:

    100a +10b + c -a -b -c = 99a - 9b = 9(9a -b)
     
  4. Aug 11, 2011 #3

    gb7nash

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    Phinds meant 9(11a -b), but the major point is that this is a multiple of 9.
     
  5. Aug 11, 2011 #4

    phinds

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    HA ... it was so simple, I screwed it up. :tongue:

    Actually, my point was not that this is a multiple of 9, that's trivial. My point is that if you understand positional notion then most such "problems" are trivial.
     
  6. Aug 11, 2011 #5
    You may be interested to know that there are a wide variety of divisibility tests which are easily proven using positional notation and modular arithmetic as phinds has mentioned. This particular question is a consequence of the divisibility test for 9:

    A number is divisible by 9 if and only if the sum of its digits is.

    You can search for similar divisibility tests, the test for 11 is quite interesting for example,

    A number is divisible by 11 if and only if the alternating sum of its digits is.

    So take for example 2816 and add the digits in alternating order 2 - 8 + 1 - 6 = -11 which is divisible by 11. So that means the original number is also.
     
  7. Sep 9, 2011 #6
    I have just stumbled upon something fascinating about addition, subtraction, multiplication and division of digitsums and was searching the net to see if someone else has already discovered this. The best I could find on this topic, was this forum, so maybe someone here can tell me, if this is indeed a new discovery:

    I discovered that the most simplified (i.e. single digit) digitsum of any number obtained through summation, will ALWAYS be equal to the sum of the simplified digitsums of the individual terms. For example 15+37+26 = 78. The digitsum of 78 is 15 (7+8) and the digitsum of 15 is 6 (1+5).

    Now for the individual terms: 6 (digitsum of 15) + 1 (digitsum of 10) + 8 (digitsum of 26) = 15, of which the digitsum is again 6 ! This holds for ANY summation, regardless of the number of digits per term.

    The second discovery was that the same holds true for subtraction multiplication and division. With multiplication for example, the digitsums of the terms are just multiplied instead of added. For example 15x37x26=14430. The digitsum of 14430 is 12 and 1+2 is 3.

    Now 6x10x8=480, of which the digitsum is also 12 with 1+2 again being 3 !

    The only practical use for this phenomenon would have been as a form of "parity" checking in the days before calculators. For example in the case of a lenghty addition sum, if the digitsum of your end result did not match the sum of the digitsums of all the individual terms, it would have meant that your answer was faulty.

    It would be nice if someone could come up with a formula explaining this phenomenon.
     
  8. Sep 9, 2011 #7

    HallsofIvy

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    That's called "casting out nines" and works for exactly the reason phinds explained in post #2.
     
  9. Sep 9, 2011 #8
    Cool, thanks, now I know what it is called. Should have known there's nothing new under the sun. If only I had lived a few centuries earlier :) Personally I think my method is a much simplified version though, although it regrettably obscures the mechanics behind it.
     
  10. Oct 31, 2013 #9
    I found an answer!
    I have the explanation in at the bottom of another thread. https://www.physicsforums.com/showthread.php?p=4315410#post4315410

    The formula for finding the sum of the digits of a number, x, is http://latex.codecogs.com/gif.latex?x%20-%209\sum_{n=1}^{\infty}{\left%20\lfloor%20\frac{x}{10^n}%20\right%20\rfloor}

    which is pretty much "x - 9*(maths)"
    When you take x minus that, you're just left with 9*(maths).
     
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