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Necessary But Not Sufficient Check For Addition

  1. Nov 21, 2011 #1
    When I was in grade school my father told me about a method of checking my addition when I added columns of numbers. I hold degrees in mechanical engineering but none of my math courses ever broached this subject as it is pretty much useless for engineering type problems.

    Suppose you added three numbers, say 389+248+730. Adding correctly one gets 1367. Now let's add the digits of each number.

    3+8+9=20
    2+4+8=14
    7+3+0=10

    Now let's add the two digits of each of the previous sums.

    2+0=2
    1+4=5
    1+0=1

    Adding the result one gets

    2+5+1=8

    Going back to the original sum of 1367 and adding digits one gets

    1+3+6+7=17

    Adding the remaining two digits one gets

    1+7=8

    The fact that all this addition of digits produces the same final digit is a necessary condition that the addition was performed correctly.

    Can someone please explain to me why this is so.

    It works for any amount of numbers.
     
  2. jcsd
  3. Nov 21, 2011 #2
    Hi, LawrenceC,
    the trick has a postdata: if you change *any* (or all) of the digits 9's by 0's, or any (or all) of the 0's by 9's, the result is the same. If fact, you can add as many arbitrary 9's as you please at any place, without changing anything: 9 appears to behave like a 0, to all effects.

    The method is based on the fact that 10 = 9 + 1; in other words, when dividing 10 by 9, it leaves a residue of 1. Residues have an arithmetic of their own: if you were to add (or multiply, for that matter) 389+248+730, and were only interested in what remains after dividing by 9 and not in the whole result, you could add just the residues of the participating numbers in order to get the residue of the result.

    A number like 389, representing 3 hundreds, 8 tens and 9 units, is an arithmetic operation waiting to happen: 3x10x10 + 8x10 + 9. Now, the residue modulo 9 of the result of that operation can be found, as mentioned, by operating with the residues of the participating numbers... and so all 10's can become 1's: the residue is the same as in the operation 3x1x1 + 8x1 + 9, which is just the sum of the digits 3 + 8 + 9.

    See the Wikipedia page for "http://en.wikipedia.org/wiki/Casting_out_nines" [Broken]", and ask for more if needed.
     
    Last edited by a moderator: May 5, 2017
  4. Nov 21, 2011 #3

    micromass

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    The trick is that you are calculating the remainder modulo 9.

    Let's take x and y integers. We say "x=y (mod 9)" if and only if 9 divides x-y. For example, you have

    [tex]0=9=18=27~(mod~9)~\text{and}~34=7~(mod~9)[/tex]

    Now let [itex]a_na_{n-1}...a_2a_1a_0[/itex] be an integer. Of course we can write

    [tex]a_na_{n-1}...a_2a_1a_0=a_n*10^n+a_{n-1}*10^{n-1}+...+a_2*10^2+a_1*10+a_0[/tex]

    Let's calculate this (mod 9). It is obvious that [itex]10^n=1[/itex] (mod 9). Indeed: [itex]10^n-1=9999...9[/itex] which is divisible by 9. Thus

    [tex]a_n*10^n+a_{n-1}*10^{n-1}+...+a_2*10^2+a_1*10+a_0 = a_n+a_{n-1}+...+a_2+a_1+a_0[/tex]

    Now what you say follows immediately. Indeed, if x and y are integers and if x=x' and y=y' (mod 9), then x+y=x'+y' (mod 9).

    So for example, 54 + 34 = 0+7=7 (mod 9). But on the other hand 54+34=88=7 (mod 9). The trick is of course that the sum of the digits is equal to the original number in (mod 9).
     
  5. Nov 21, 2011 #4
    Thanks for the explanations. It has cleared the matter up for me.
     
  6. Nov 21, 2011 #5
    and it works with multiplications too.
    13*7=91 but here you don't add 13 to 7 but you mulitply 4 ( coming from 1+3 = 4 from 13 ) by 7 to get 28. Now you add the digits of the result 2+8 and it is of course equal to 9+1. But if by mistake you switch digits the sum of digits is not helpful because 13*7=19 will work too for the sum of digits.
     
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