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Casting Out Nines and Elevens


Date: 09/19/97 at 00:45:44
From: Anonymous
Subject: Throw nine away

At a parent-teacher meeting this evening, the teacher asked the 
parents why nine is used in proving a math answer. She did not know 
the answer, and the parents didn't either. Can you help? Our fourth-
grade students will be taking this new method of problem solving.


  EXAMPLE    6313    6+3 = 9 throw away   1+3 = 4
                     1452    4+5 = 9 throw away   1+2 = 3
                     7765    7

  If you're left with 7, then your answer is correct.

Thank you,

Susan Racela


Date: 09/25/97 at 11:49:31
From: Doctor Rob
Subject: Re: Throw nine away

This procedure for checking arithmetic is called "casting out nines,"
and has been known for some centuries.  It is based on modular 
arithmetic with modulus nine.

Nine is used because it is the largest integer such that it fits into 
the following pattern:

         10 = 9*    1 + 1
        100 = 9*   11 + 1
       1000 = 9*  111 + 1
      10000 = 9* 1111 + 1
     100000 = 9*11111 + 1
        ...       ...

The powers of 10 on the left represent the place values of the various
digits. The 1 on the far right represents the digit itself. Your 
example:

       6313 = 6*1000 + 3*100 + 1*10 + 3
            = 6*(9*111+1) + 3*(9*11+1) + 1*(9*1+1) + 3
            = 9*(6*111+3*11+1*1) + 6 + 3 + 1 + 3
       6313 - (6+3+1+3) = 9*(6*111+3*11+1*1)

This shows that the number and the sum of its digits differ by a 
multiple of nine.  This is true for every counting number (positive 
integer). A consequence of this is that if, in an addition, 
subtraction, or multiplication problem, you replace each number by the 
sum of its digits, you change the result by a multiple of nine.

The usual notation for this situation is to say that two numbers 
x and y are congruent modulo m if their difference is a multiple of m.  
This is written x = y (mod m).

In the case at hand 6313 = 6+3+1+3 = 13 (mod 9), and further
13 = 1+3 = 4 (mod 9). Similarly 1452 = 1+4+5+2 = 12 (mod 9), and 
further 12 = 1+2 = 3 (mod 9). The answer 7765 = 7+7+6+5 = 25 = 2+5 = 
7 (mod 9).

If the answer 7765 is correct, then 6313 + 1452 = 4 + 3 = 7 (mod 9) 
and 7765 = 7 (mod 9) provides a degree of checking. If these two 
numbers disagree, you know you have made an arithmetic error. If they 
are the same, you have a certain degree of confidence that you have 
not, although not certainty!  For example, a transposition of digits 
(7675, say) cannot be detected by casting out nines.

There is a related technique called "casting out elevens" which is 
based on the following pattern:

            1 = 11*     0 + 1
           10 = 11*     1 - 1
          100 = 11*     9 + 1
         1000 = 11*    91 - 1
        10000 = 11*   909 + 1
       100000 = 11*  9091 - 1
      1000000 = 11* 90909 + 1
     10000000 = 11*909091 - 1
          ...         ...

     6313 = 6*1000 + 3*100 + 1*10 + 3
          = 6*(11*91-1) + 3*(11*9+1) + 1*(11*1 - 1) + 3
          = 11*(6*91+3*9+1*1) - 6 + 3 - 1 + 3
     6313 - (3-1+3-6) = 11*(6*91+3*9+1*1)

It uses the alternating-sign sum of digits, working *right to left*:

     6313 = 3 - 1 + 3 - 6 = -1 = 10 (mod 11)
     1452 = 2 - 5 + 4 - 1 =       0 (mod 11)
     7765 = 5 - 6 + 7 - 7 = -1 = 10 (mod 11)

The fact that 10 + 0 = 10 (mod 11) gives an additional check on the
validity of the arithmetic.

Casting out elevens is much less well known than casting out nines.  
It WILL detect digit transpositions.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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