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Casting Out Nines for 2nd Graders

Date: 07/05/2004 at 20:39:42
From: Donna
Subject: Casting Out Nines

Dear Dr. Math,

Can you please send me a simpler explanation of WHY the Casting Out
Nines method works than you have in your archives?  I understand how
to do it, but I need to know why it works.  I have to present this
method so that second graders can understand why it works.

Date: 07/05/2004 at 22:58:49
From: Doctor Peterson
Subject: Re: Casting Out Nines

Hi, Donna.

I've tried a number of times to find a good way to prove the method 
at an elementary level, and found that any really clear explanation 
requires some knowledge of modular arithmetic and algebra.  Without 
those ideas, there's just too much work needed to work around them.

But that's talking mostly about proofs.  Second graders wouldn't 
appreciate an actual proof anyway, so we can just look for a 
plausibile explanation to show HOW it works -- what's going on 
behind the scenes.  That may be easier to handle, though we'll have 
to keep it extremely simple.

The basic idea, leaving out all the details, is this:  Adding the 
digits of a number decreases it by a multiple of 9, so repeating the 
process until you have a single digit leaves you with the remainder 
after division by 9 (or with 9 if that remainder is 0).  (In advanced 
terms, the digit sum is congruent to the original number modulo 9). 
When numbers are added or multiplied, the remainder of the result is 
the same as the sum or product of the remainders of the given 
numbers.  (In advanced terms, the sum and product are well-defined in 
modular arithmetic; that is, they preserve congruence.)  So if the 
remainders on both sides of an equation are not equal, then neither 
are the values of both sides themselves.

For second graders, we'll want to work with a specific example rather 
than with generalities, and with addition rather than multiplication.  
Take this sum as our example:

  24 + 37 = 51

Is this correct?  Well, we look at the left side, the sum.  We add the 
digits of 24 and get 6; we add the digits of 37 and get 10, then add 
again to get 1.  Now we add 6 and 1 to get 7.  On the right side, we 
add the digits of 51 to get 6; since this is not equal to 7, our 
answer is wrong.  (The right answer is 61, which does give a digit 
sum of 7.)

Now, what is happening when we do that?

Instead of adding 24 + 37, we're adding 6 + 1.  Now, 6 is 18 less than 
24, and 1 is 36 less than 37, so their sum is 18 + 36 less than the 
real sum.  Do you see that 18 and 36 are both multiples of 9?  That 
means that the sum we get, 7, is some number of 9's less than the real 
sum.  This will always happen.  [Why? Because replacing 20+4 with 2+4 
takes away 20 and adds 2, which is the same as taking away 10 and 
adding 1 twice.  But that means taking away a multiple of 9.]

When we do the same thing to the 51, the sum of the digits, 5+1=6, is 
45 less than 51 itself; so again we get a number that is some number 
of 9's less than the actual number.

But that means that the numbers we get on the two sides should 
themselves be a multiple of 9 apart.  In fact, since they are both 
single-digit numbers, they should be equal (unless one is 0 and the 
other is 9).  Since they are not, the numbers can't be right.

I'm not at all sure this will work for the average second grader; even 
though I avoided going into detail about place value and remainders, I 
had to bring in multiples of 9.  It's hard to avoid something like 
that, since that is what the technique is all about under the hood:  
multiples, remainders, modular arithmetic.

Please let me know if you find a nice way to express these ideas for 
that age.

- Doctor Peterson, The Math Forum 

Date: 07/06/2004 at 19:10:02
From: Donna
Subject: Thank you (Casting Out Nines)

Thank you very much for the speedy reply.  My presentation is 
tomorrow.  Maybe the other educators in my group will have a 
suggestion for explaining it to second graders.  I'll let you know if
they do.  Again, thank you!!!
Associated Topics:
High School Number Theory

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