# Digit Sums and "casting out 9's" - do what?? Please explain . . .

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I honestly didn't think 7th grade math would prove so annoying! LOL I'm working through CLE 702 with my dd in attempts to get her back up to speed with her math before 8th grade since Saxon 8/7 totally traumatized her earlier this year. :crying:

Today our lesson is on "digit sums" - now, either I don't recall being taught this, or I wasn't. DD confirms her private school teacher skipped this lesson in 5th when it was introduced saying: "no one ever uses this method - it's confusing!" Umm, yes, OK!

Can someone direct me to a "digit sums" for dummies you tube? I've tried Kahn Academy with no luck. I can show her how to do it with the casting out 9's - what I can't seem to figure out (and neither can she) is WHY we would do this? What useful purpose does it serve?

I also can't for the life of me figure out HOW they feel like the number 548 has a digit sum of 8. When I add those numbers together, I get 5 + 4 + 8 = 17 - nothing like the "8" I get if I "cast out 9's" *sigh* Maybe my problem is with the word "sum" - I thought "sum" meant add the numbers together. Honestly, how is it possible that I made it through AP Cal and placed out of math in college but I never, ever, learned this?! (Unless DD's math teacher was right and it is confusing/irrelevant!) ;)

When I google it, I get things like this:

To find the digit sum of a number you plus the first 2 numbers together, example:

the number 18, 18 is made up of 2 digits 1 and 8.

If you add the 2 digits together you get a total of 9.

So that is the digit sum of 18.

But if you use a higher number like 66 then the method has a small change.

First if you add the two digits together it becomes 12.

But 12 isn't a single digit sum.

So again the 2 digits have to be added together to make a single digit sum.

1+2=3 so 3 is the single digit sum for the number 66.

Umm, what on earth?? My brain hurts. :huh:

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I have never heard of doing what your math program is asking, but based on what you posted, 548's digits add up to 17 and then you continue on to add 17's digits and get 8.

The only reason I have ever learned for adding up digits is to determine if a number is divisible by 3.

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Casting out 9s is evil :laugh: We ran into it with A Beka Math. To be honest we came to the desicion to skip all reference to the idea. My husband teaches in public school and had never heard of it. I have a teaching degree and never heard of it. No one uses it. I think it is one of the 'cutesy' things someone came up with to take short cuts but has no real world application.

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I honestly didn't think 7th grade math would prove so annoying! LOL I'm working through CLE 702 with my dd in attempts to get her back up to speed with her math before 8th grade since Saxon 8/7 totally traumatized her earlier this year. :crying:

Today our lesson is on "digit sums" - now, either I don't recall being taught this, or I wasn't. DD confirms her private school teacher skipped this lesson in 5th when it was introduced saying: "no one ever uses this method - it's confusing!" Umm, yes, OK!

Can someone direct me to a "digit sums" for dummies you tube? I've tried Kahn Academy with no luck. I can show her how to do it with the casting out 9's - what I can't seem to figure out (and neither can she) is WHY we would do this? What useful purpose does it serve?

I also can't for the life of me figure out HOW they feel like the number 548 has a digit sum of 8. When I add those numbers together, I get 5 + 4 + 8 = 17 - nothing like the "8" I get if I "cast out 9's" *sigh* Maybe my problem is with the word "sum" - I thought "sum" meant add the numbers together. Honestly, how is it possible that I made it through AP Cal and placed out of math in college but I never, ever, learned this?! (Unless DD's math teacher was right and it is confusing/irrelevant!) ;)

When I google it, I get things like this:

Umm, what on earth?? My brain hurts. :huh:

You keep adding the digits together until you get a single digit number. The easiest way is to discard the numbers that add together to make 9 (like in your examle of 548, 5+4=9, so you cross that out and you are left with 8). Otherwise, just keep adding the digits together until you get a single digit number (5+4+8=17, 1+7=8). You will get the same number either way.

Do this both vertically and horizontally with an addition problem, and your digit sums should be the same. If not, there is an error in your answer. It is a quick, simple check to see if you solved an arithmetic problem correctly.

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Let me give you an example:

274

+639

I added it together and got a sum of 913.

To check my answer, I first find the sum of both addends. The top row is 2+7+4. I know that 2+7 makes 9, so I can discard those numbers, which leaves me with 4. The second row is 6+3+9. I can discard 6+3, because that makes 9, and I can also discard the 9, which leaves me with a total of zero. I add the sums of both addends together (4+0), which equals 4.

My second step is to add together the digits of my answer (9+1+3). I can discard the 9, leaving me with 1+3=4. Since I came up with same digit sum for the addends and for the answer (4), chances are that my answer is correct.

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While it can be used for checking sums (which is what CLE introduces it for), it's especially useful for divisibility rules. For example:

Divisible by 3?

12345 = 1x10000 + 2x1000 + 3x100 + 4x10 + 5

= 1 + 1x9999 + 2 + 2x999 + 3 + 3x99 + 4 + 4x9 + 5

Now, take out all the numbers being multiplied by 9 or 99 or 999, etc. - you know those are definitely divisible by 3.

You're left with: 1 + 2 + 3 + 4 + 5 = the digit sum. If THAT sum is divisible by 3, then the original number is divisible by 3. That sum is 6 (15's digit sum is 6), and 6 is divisible by 3, so the whole number is divisible by 3.

The same thing works for 6 and 9 (for 6, if it's divisible by 3 and even, it's divisible by 6... for 9, if the digit sum is divisible by 9, the whole number is divisible by 9).

But yeah, the way CLE uses it:

..12345 -> 6

+12345 -> 6

_______

..24690 -> 21 -> 3

6+6 = 12 -> 3

The two digit sums are equal, so our math was probably correct. If I had forgotten to regroup and said it was 24680, by digit sum would be 2, and we'd know that's a wrong answer for sure. Neat little trick, but I'm not sure how much time it actually saves. You might catch your answer just as quickly by looking over the steps of each addition.

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Let me give you an example:

274

+639

I added it together and got a sum of 913.

To check my answer, I first find the sum of both addends. The top row is 2+7+4. I know that 2+7 makes 9, so I can discard those numbers, which leaves me with 4. The second row is 6+3+9. I can discard 6+3, because that makes 9, and I can also discard the 9, which leaves me with a total of zero. I add the sums of both addends together (4+0), which equals 4.

My second step is to add together the digits of my answer (9+1+3). I can discard the 9, leaving me with 1+3=4. Since I came up with same digit sum for the addends and for the answer (4), chances are that my answer is correct.

Thanks!

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CLE emphasizes checking your own work quite a bit. Using digit sums is just one way of doing that. As previously stated, casting out nines is just a shorter way of doing digit sums. I do find the digit sum method very effective when checking multiplication problems especially. I think the ubiquity of calculators in schools for checking work has reduced the prevalence of this technique but it is very fast and handy.

For example,

9791 x 813 = 7960083

Take the digit sum for 9791 => (8)

• w/o casting nines = 9+7+9+1 = 26 = > 2+6= 8
• using casting out - ignore 9s and just add 7+1 = 8

Take the digit sum for 813 => (3)

• w/o casting nines = 8+1+3=12 => 1+2 = 3
• using casting out - ignore 8 and 1 since they = 9 and you have 3

Finally,take the digit sum for the product 7960083 =>. (6)

• w/o casting nines = 7+9+6+8+3 = 33 => 3+3 = 6
• using casting out - ignore 9 as well as 6 and 3 since they sum to 9, take 7+8 = 15 => 1+5 = 6

To check the problem, multiply the digit sums for the 2 factors (8 x 3). The digit sum of their product (24 => 2+4= 6) should be equal to the digit sum of the actual product.

This may seem laborious at first but if you just practice a little esp. with the casting nines, it is a very quick method for checking. Literally, just taking seconds and certainly much quicker than remulitplying everything out or checking by division.

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Using digit sums is especially important for people who plan to own small businesses and need to do their own accounting without the aid of calculators or other technology.

It is a neat trick but not necessary for higher level math.

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I don't remember learning it in school either, but R&S brings it in in the 5th grade book and again in our 6th grade book. I really like it. It is actually a quick way to check your answers in multiplication or long division problems. My dd knows her x tables and such, but often makes careless mistakes in her bigger problems. Learning to check her work by casting out nines is a quick way to check her answers on tests. If it doesn't come out right, then she needs to redo the problem. If it does, then she doesn't have to do it twice.

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Thanks everyone! We had an 'aha!' moment when we used it to check a sum later in the lesson. Both of us were totally shocked at how it could be used! So I guess we both learned something today! lol I think I was hung-up on the use of the word 'sum' because I couldn't reconcile that with throwing numbers out or adding together the product etc. Ignoring all that, it's a really cool trick!

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*I* thought it was a neat thing when I learned it (CLE 400, Abeka teaches it in 4th or 5th, too). I used it mostly with division (3s/9s, & 6s if you check 3s & the # is also even) as boscopop pointed out.

DD#1 needs a reminder on this one as she makes a lot of "silly" mistakes in her work & could definitely use it to check her work on tests, although it wouldn't help her with percentage/place value issues.

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It's a cool trick, but if it stresses the kid out, you can skip it. NONE of my kids have learned this. Eldest is a math major. It's all good.

I always thought it would be fun to save these for a special "math tricks" unit maybe done in summer or over weekends etc. Maybe.

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We just did this is CLE 500 and it drove me nuts trying to figure out what it was and what it meant. I also checked on Khan Academy and couldn't find anything. In the end, there was a good example and explaination in the student book for CLE but not the teacher's book.

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You can also use it for numbers that are too long for a calculator screen, and not just for addition.

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Thanks to whomever posted the answer to the OP.

:confused1: When I heard "casting out 9s" it sounded like the nines had done something wrong and were being cast out. I envisioned someone with a scarlet 9 on their dress :laugh:

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I'm not all that familiar with this, though I've heard of this term before.

I think you would use this method to check your arithmetic. Like if you get the same digit some across and downwards then your answer is right...

When you do arithmetic with larger numbers, it can be a lot faster than reversing the work to check the whole process.

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If you do CLE math from beginning on it's actually a choice of whether you want to do it or not. We're on CLE 5, I've done 4 and both levels have said that you can check your math this way or another way in the TM. This choice is just not reflected in LU. I find this way, totally confusing and annoying. So I skip it. We check it by reversing numbers or equations. (edited to add: these posts were helpful though...so now I have to try it. :))

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While it can be used for checking sums (which is what CLE introduces it for), it's especially useful for divisibility rules. For example:

Divisible by 3?

12345 = 1x10000 + 2x1000 + 3x100 + 4x10 + 5

= 1 + 1x9999 + 2 + 2x999 + 3 + 3x99 + 4 + 4x9 + 5

Now, take out all the numbers being multiplied by 9 or 99 or 999, etc. - you know those are definitely divisible by 3.

You're left with: 1 + 2 + 3 + 4 + 5 = the digit sum. If THAT sum is divisible by 3, then the original number is divisible by 3. That sum is 6 (15's digit sum is 6), and 6 is divisible by 3, so the whole number is divisible by 3.

The same thing works for 6 and 9 (for 6, if it's divisible by 3 and even, it's divisible by 6... for 9, if the digit sum is divisible by 9, the whole number is divisible by 9).

This is important. Keep it in mind for when you get to fractions, because it will make it much easier to recognize when a fraction is in simplest form or whether the numerator and denominator have factors in common.

Also, when you have to rewrite numbers as a product of prime factors (pre-algebra), being able to recognize quickly what a number is divisible by can help.

And then, proving WHY these rules work is a wonderful exercise for pre-algebra or algebra students to demonstrate their understanding and mastery of the concept of place value and of how numbers work.

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I never learned the "casting out 9s" trick for checking computation. I don't feel I missed anything.

This is important. Keep it in mind for when you get to fractions, because it will make it much easier to recognize when a fraction is in simplest form or whether the numerator and denominator have factors in common.

Also, when you have to rewrite numbers as a product of prime factors (pre-algebra), being able to recognize quickly what a number is divisible by can help.

And then, proving WHY these rules work is a wonderful exercise for pre-algebra or algebra students to demonstrate their understanding and mastery of the concept of place value and of how numbers work.

:iagree:

For fun, divisibility videos:

Divisibility Rules for 2, 4, 5, and 25

Divisibility Rules for 3 and 9

How to Tell if a Number is Prime

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We skipped the "casting out 9s" in Abeka.

We certainly added digits to determine divisibility rules of 3,6,9 (i never knew that!)... THAT was really useful!

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We ran into this being an issue for dd as well. Skipped it entirely.

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