Casting Out Nines for 2nd GradersDate: 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 http://mathforum.org/dr.math/ 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!!! |
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