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Number 9

  1. Aug 29, 2007 #1
    It seems that if the number 9 is multiplied by a number, the resulting digits of that number will sum to 9. Does this always hold true? Is there a simple proof for this, or at least a proof up to some finite number?
  2. jcsd
  3. Aug 29, 2007 #2
    I assume that by a "number" you are referring to a positive integer.

    9x11 = 99
    9+9 = 18 != 9 (although 1+8 = 9)

    It's easy to prove that sum of the digits is always an integral multiple of 9, though.
  4. Aug 29, 2007 #3
    It's called "Casting out Nines." The fact is that 10 congruent to 1, Modulo 9, (you might have to look up modulo) and thus 10^n==1 Mod 9 for n a positive integer. So that:

    [tex]\sum a_i10^n \equiv \sum a_i [/tex] Modulo 9.

    This fact was often used by bookkeepers in the old days as a check on figures. After calculating the answer, they would go through the problem again "casting out nines," to get a check.

    For example: 125x275 = 34375== 4 Mod 9. and 125==8 Mod 9, 275 ==5 Mod 9. So that 125 x 275 ==8x5=40==4 Mod 9.
    Last edited: Aug 29, 2007
  5. Aug 29, 2007 #4
    This is a consequence of the behavior of modular arithmetic, plus the fact that 10 = 1 (mod 9).

    Your multiple of 9, as any other number, can be expressed as its digits multiplied by powers of 10; say, 1845 is 1.10^3 + 8.10^2 + 4.10 + 5. Now, being a multiple of 9 means that its remainder when divided by 9 is exactly zero. So you would write, 1845 = 0 (mod 9).

    Arithmetic modulo some number n (in our case, modulo 9) has this useful properties:
    If a = a' (mod n) and b = b' (mod n), the following are true:
    • a + b = a' + b' (mod n)
    • ab = a'b' (mod n)
    • a^i = a'^i (mod n), for any integer i

    So any expression you can write using sums, products or powers to some integer, will produce the same result whether
    • you calculate the whole expression first, and then finally take the remainder modulo n, or
    • you take the remainders modulo n of your numbers, and do the operation on the remainders, taking the modulo n after each operation.

    For the example number above, you can do all operations "mod 9", taking notice that 10 = 1 (mod 9). Then you have:
    = 1.10^3 + 8.10^2 + 4.10 + 5
    = 1.1^3 + 8.1^2 + 4.1 + 5
    = 1 + 8 + 4 + 5 (mod 9)​
  6. Aug 29, 2007 #5
    Another statement of what has been explained Robort Ihnot and Dodo is that whenever you add nine it corresponds to adding ten and removing one. In other words, you move digit values from the tens to the ones, but you do not change their sum.
  7. Aug 29, 2007 #6


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    Can you run through this a little slower? This seems like an incredibly powerful tool for checking numbers (for those of us old fools who still try to do it without computers).

    [ EDIT ] Never mind, a quick Wiki did the trick. I've got to try that next time I have several large numbers!

    MAN I wish I'd known that trick in grade school and high school! Many of my tests and assignments were lower than expected because of no expedient way of cross-checking.
    Last edited: Aug 29, 2007
  8. Aug 29, 2007 #7
    The biggest use of "casting out nines" was in Accounting, where it might be discussed. Of course, it is not a fool-proof test, but serves as a quick check. A common error, so I was told, is for the writer to reverse digits, such that 23 was put down as 32. Of course casting out nines will not help with that error.

    I have noticed that few students today even understand that to multiply 9 by a single digit number, from 2-9, you subtract 1 and add what it takes to make nine for the second digit. Thus 9x5 = 45. This is such a simple thing if you are learning the multiplication tables.
    Last edited: Aug 29, 2007
  9. Aug 29, 2007 #8


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    In base 8, 7 has the same property as 9 does in base 10. Similar with 0xF in base 16.
  10. Aug 29, 2007 #9

    Chris Hillman

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  11. Aug 29, 2007 #10


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    Hmmm...I think this sentence could have been better constructed, wot?
  12. Aug 30, 2007 #11

    Gib Z

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