Casting Out Nines to Check Arithmetic

Date: 08/26/99 at 19:29:17
From: Jason Fowler
Subject: Division, Multiplication, Subtraction, and Addition

My teacher was talking about casting out 9's. She said it was the 
easiest way to check the problems. No one in the class understands 
this method. She is a new teacher, we're in the 6th grade, and she was 
teaching high school math.

Thank you,
Jason Fowler

Date: 08/27/99 at 08:56:02
From: Doctor Peterson
Subject: Re: Division, Multiplication, Subtraction, and Addition

Hi, Jason.

Casting out nines is not high-school math if all you want to do is use 
it; but it can take some effort to explain why it works without 
getting into hard stuff. Here are two explanations of it in our 
archives, both of which get into some deeper ideas than you want by 
way of explaining how it works:

   Casting Out Nines and Elevens
   http://mathforum.org/library/drmath/view/55805.html

   Casting Out Nines   
   http://mathforum.org/library/drmath/view/55831.html  

I'll give you a quick explanation of how to do it, without the big 
words.

First, for any number we can get a single digit, which I will call the 
"check digit," by repeatedly adding the digits. That is, we add the 
digits of the number, then if there is more than one digit in the 
result we add its digits, and so on until there is only one digit 
left. 

For example, for the numbers 6395 and 1259, we get:

     6395 --> 6 + 3 + 9 + 5 = 23 --> 2 + 3 = 5
     1259 --> 1 + 2 + 5 + 9 = 17 --> 1 + 7 = 8

Now, it turns out that if you add or multiply a set of numbers, the 
check digit of their sum is the same as the check digit of the sum of 
the check digits. You can think of it like this:

     numbers -----> digits
        |    check    |
        |     (1)     |add
        |             |(2)
        |             V
        |add         sum -----> digit
        |(4)             check    |
        |                 (3)     |equal?
        |         (5)             | (6)
        V        check            V
       sum -------------------> digit

That is, if you calculate the check digit (1) of each number you're 
adding, then add these (2) and calculate the check digit of the sum 
(3), that should be the same as the check digit (5) of the sum (4) you 
are checking.

In our example, the sum of the numbers is

       6395
     + 1259
     ------
       7654 --> 7 + 6 + 5 + 4 = 22 --> 2 + 2 = 4

with check digit 4, and the sum of their check digits is

       5
     + 8
     ---
      13 --> 1 + 3 = 4

So the check digit of the sum is 4, and the check digit of the sum of 
the check digits 5 and 8 is also 4. If they didn't agree, we'd know 
something was wrong. Here's my diagram:

     6395 -------> 5
     1259 -------> 8
       |   check   |
       |           |add
       |           |
       |           V
       |add       13 -------> 4
       |              check   |
       |                      |equal? yes!
       |                      |
       V         check        V
     7654 ------------------> 4

Similarly, for multiplication, the product of the numbers is

        6395
     x  1259
     -------
     8051305 --> 8 + 0 + 5 + 1 + 3 + 0 + 5 = 22 --> 2 + 2 = 4

and the product of the check digits is

       5
     x 8
     ---
      40 --> 4 + 0 = 4

which agrees with our product:

      6395 -------> 5
      1259 -------> 8
        |   check   |
        |           |mult
        |           |
        |           V
        |mult      40 -------> 4
        |              check   |
        |                      |equal? yes!
        |                      |
        V         check        V
     8051305 ----------------> 4

(You wouldn't normally get the same check digit for the result of the 
sum and the products; I just picked a weird example.)

You can also apply the process to subtraction and division, but 
because of some special cases you have to deal with, it's easier to 
transform the problem to an addition or multiplication. For example, 
to check the subtraction:

       7654
     - 6395
     ------
       1259

you would transform it to the addition I did above. To check this 
division:

          _____6395_rem 342
     1259 ) 8051647

you would transform it to the multiplication I did above, by adding 
the check digit of the remainder to the product of the check digits of 
the quotient and the divisor, and checking whether this is equal to 
the dividend:

        6395 -->   5
     x  1259 --> x 8
     -------     ---
     8051305      40 -->   4
     +   342 ----------> + 9
     -------             ---
     8051647 ---> 4 <---  13

In other words, you apply casting out nines not to the subtraction or 
division itself, but to the standard check, in which you reverse the 
problem by adding the subtrahend to the difference or multiplying the 
quotient by the divisor.

If the check digits don't come out right, you must have made a mistake 
in your arithmetic (either in the problem you're checking, or in 
calculating the check digit); but if the check digits agree, your 
work could still be wrong, such as if you switched two digits when you 
were copying. In fact, I use a variant of this method when I balance 
my checkbook. If I get the wrong balance, I know my calculator didn't 
add wrong, but I may have entered something wrong. If the check digit 
for my balance is the same as what the bank says, I can guess that I 
reversed two digits somewhere; if they are different, it's more likely 
that I dropped a digit, or perhaps a whole transaction.

If you want to know what this has to do with nines, or why it works, 
check out the other answers I referred to above. The basic idea is 
that the check digit is essentially the remainder after you divide by 
9. (A slightly more advanced way of working with these check digits is 
to treat a result of 9 as a zero, so that check digits are always 
between 0 and 8 rather than 1 and 9, making it a genuine remainder.)

You may notice that when you add the digits of 6395, if you just 
ignore the 9, and the 6+3 = 9, you still end up with 5 as your check 
digit. This is because any 9's make no difference in the result. 
That's why the process is called "casting out" nines. Also, at any 
step in the process, you can add digits, not just at the end: to do 
8051647, I can say 8 + 5 = 13, which gives 4; plus 1 is 5, plus 6 is 
11, which gives 2, plus 4 is 6, plus 7 is 13 which gives 4. I never 
have to work with numbers bigger than 18.

I hope this clarifies what you're doing. It takes a lot of words to 
explain, but it's really easy to do. Keep at it and you'll get the 
idea. If you want a simpler explanation of WHY the method works than 
we have in our archives, write back and I can send that to you too.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   

Date: 03/23/2003 at 13:28:21
From: Sarah Dodge
Subject: Casting out nines

Dear Dr. Math, 

I don't understand casting out nines. Can you help me?

Here is an example
  3942
- 1581
-------
I know that it equals  2361

I think that you would add the numbers together and you would throw 
out the nines, so it would be
  3942 (0)
- 1581 (6)
so what wwould the answer be?  (3?)

How do you get that though ?

Date: 03/24/2003 at 12:40:37
From: Doctor Peterson
Subject: Re: Casting out nines

Hi, Sarah.

I would usually check a subtraction by checking the equivalent 
addition; here 2361 + 1581 = 3942. In this case, the reduced digit 
sums are 3, 6, and 0, which is correct since 3+6=9 which reduces to 0.

But you can cast out nines to check subtraction directly, if you add 
one step. In this example you want to subtract 6 from 0. To make the 
0 big enough to subtract, you can "borrow" a 9, by adding 9 to the 
zero. (Just as you reduce by casting out nines, you can grab an extra 
nine when you need one.) Now you are subtracting 6 from 9, which gives 
you the 3 you expect.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/