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# Beast 5B question

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Hello Hive, My DD is in tears this morning because she of a concept in Beast 5B (workbook page 42) that she cannot understand.  The text states:

"The GCF of any 2 numbers is the same as the GCF of either number and their difference." (GCF - greatest common factor).

She understands if I give her a bunch of sample numbers and show her how it works, but she's having a hard time understanding it intuitively, and I'm at a loss for how to explain it an intuitive way. I was choosing sample numbers and drawing number lines to show her skip counting patterns, they way the Beast text does. Do any of you have alternative ways of visualizing or explaining this concept?

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Hello Hive, My DD is in tears this morning because she of a concept in Beast 5B (workbook page 42) that she cannot understand.  The text states:

"The GCF of any 2 numbers is the same as the GCF of either number and their difference." (GCF - greatest common factor).

She understands if I give her a bunch of sample numbers and show her how it works, but she's having a hard time understanding it intuitively, and I'm at a loss for how to explain it an intuitive way. I was choosing sample numbers and drawing number lines to show her skip counting patterns, they way the Beast text does. Do any of you have alternative ways of visualizing or explaining this concept?

1) That seems like a really clunky definition. Are you sure that's what it says? Perhaps I'm slow, but that seems like a typo. I don't know what that underlined bit means in this context and I know what a GCF is and how to find it.

The GCF is the BIGGEST (greatest) factor that two numbers have in common. It's practically a verbal/vocabulary comprehension thing to someone who understands how to factor numbers. It's "intuitive" through understanding that terminology and seeing the connection to the math, but not really a math concept to be intuited in and of its self if that makes sense.

To find the GCF of two numbers, factor both numbers and see what is the BIGGEST factor that they have in common. That they share.

That part that says "and their difference" is just...REALLY confusing me. Do they mean difference, mathematically, like...subtract them?

OP, would you mind posting a picture of that page/definition? Perhaps they are teaching it some way that makes the extra verbiage necessary, but now I'm painfully curious to see this page myself.

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Assuming  I understand this correctly. Given two numbers x and y, there always is some gcf and you can rewrite both as a multiple of that gcf and the remaining parts k1 and k2 are relatively prime (you can think of them as a product of primes themselves if that makes it easier).

i..e   x = k1 * gcf  and y = k2 * gcf

Then when subtract the two the distributive law says the difference will still be a multiple of the gcf.

i.e. assume x is greater than y

x - y = k1 * gcf - k2 * gcf = (k1 - k2) * gcf

So this difference definitely shares the original gcf  as a common factor with both original numbers x and  y. And the gcf must still be the greatest common factor for the inverse reason. If the other part of the product k1 - k2 contained another factor > 1 with either number then this factor would have to have been in  both originals via the distributive law and we'd have a contradiction with the original gcf  being the greatest factor.

This looks a bit formal on rereading so I hope this helps.

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That part that says "and their difference" is just...REALLY confusing me. Do they mean difference, mathematically, like...subtract them?

Disclaimer - I don't have any BA. But, if you take two numbers, say, 57 and 69, their GCF is 3. And if you do 69-57, the difference is 12, and the GCF of 12 and 57 (or 69) is still 3. You can try it with other numbers too, if you want - seaben is right, but I thought I'd throw out some actual numbers to illustrate this.

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Would the quotient remainder theorem be overkill at this point, do you think? It's just that I think the Euclidean algorithm might actually really help her understand what's going on here.

Edited by Pegs

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Assuming  I understand this correctly. Given two numbers x and y, there always is some gcf and you can rewrite both as a multiple of that gcf and the remaining parts k1 and k2 are relatively prime (you can think of them as a product of primes themselves if that makes it easier).

i..e   x = k1 * gcf  and y = k2 * gcf

Then when subtract the two the distributive law says the difference will still be a multiple of the gcf.

i.e. assume x is greater than y

x - y = k1 * gcf - k2 * gcf = (k1 - k2) * gcf

So this difference definitely shares the original gcf  as a common factor with both original numbers x and  y. And the gcf must still be the greatest common factor for the inverse reason. If the other part of the product k1 - k2 contained another factor > 1 with either number then this factor would have to have been in  both originals via the distributive law and we'd have a contradiction with the original gcf  being the greatest factor.

This looks a bit formal on rereading so I hope this helps.

This is the reasoning (and yes you need both directions). Basically if d divides any two out of x, y, x-y, then it divides the third.

Would the quotient remainder theorem be overkill at this point, do you think? It's just that I think the Euclidean algorithm might actually really help her understand what's going on here.

Actually it is leading towards the Euclidean algorithm.

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Actually it is leading towards the Euclidean algorithm.

Ah! Sorry, I don't have level 5 of Beast so I wasn't aware of the context.

OP, you might consider proceeding to the EA then cycling back through the GCF problems?

Edited by Pegs

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Assuming  I understand this correctly. Given two numbers x and y, there always is some gcf and you can rewrite both as a multiple of that gcf and the remaining parts k1 and k2 are relatively prime (you can think of them as a product of primes themselves if that makes it easier).

i..e   x = k1 * gcf  and y = k2 * gcf

Then when subtract the two the distributive law says the difference will still be a multiple of the gcf.

i.e. assume x is greater than y

x - y = k1 * gcf - k2 * gcf = (k1 - k2) * gcf

So this difference definitely shares the original gcf  as a common factor with both original numbers x and  y. And the gcf must still be the greatest common factor for the inverse reason. If the other part of the product k1 - k2 contained another factor > 1 with either number then this factor would have to have been in  both originals via the distributive law and we'd have a contradiction with the original gcf  being the greatest factor.

This looks a bit formal on rereading so I hope this helps.

This makes perfect sense. But for some reason, in my mind, that English explanation doesn't communicate this. The and their difference seems....vague? I don't know.

Also, this seems like a weird way to explain it in words. The mathematical notation is very "duh" too me, but this wording? Huh. I'm thinking a little too hard.

"The GCF of any 2 numbers is the same as the GCF of either number and their difference."

I think that more than the concept, the wording of this, especially the red  and blue part is a little confusing. Maybe that is what you're daughter is struggling with OP.

re the red: There can ONLY be a greatest common factor if there are two or more.  For some reason, in my mind the word "either" would instinctively translated to one of the numbers OR the other number being factored.

12 does not have a GCF, but 12 and 16 do have a GCF.

re: the blue, I was confused because "their" seemed like it was talking about the GCF and I was all :huh: Because the GCF is just one number, so GCF-GCF would 0.

I get it now, but on the first several readings this seems like a bad English "translation" of what they are getting at. Since she's struggling with the definition that uses multiplication-terms, I would walk-through this with your daughter and find the equivalent definition that uses divisibility-terms.

That might help her see the concept without the English clouding up her math.

Basically it's saying "when you have the numbers. a, b,  then assuming that a > b, you can have: a, b and (a-b) as your 3 numbers.

If TWO of those numbers are divisible by n, then so is the third number. Guaranteed.

OP, maybe your daughter just needs help unpacking the verbiage the way that I did?

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I typed a reply last night but it never went through... thank you all for the suggestions!! I think perhaps she was just brain fried yesterday and unable to process. Today we went over it again using a number line and several sets of sample numbers. It clicked for her this time. Using the number line (sort of like using division concepts suggested by mom2bee) we saw how the GCF is essentially an interval that you can use to skip count through number alto number b. She could visualize on the line how you would have to be able to skip count through the delta (b-a) as well using the same interval. The beast workbook actually does s good job of explaining it, I think she may have just needed more examples.

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I know that no one cares but

OMG! I have been thinking about this and it just 'clicked' for me as I was going back to bed.

It's almost midnight right now but I had a eureka moment and had to come share.

I typed a reply last night but it never went through... thank you all for the suggestions!! I think perhaps she was just brain fried yesterday and unable to process. Today we went over it again using a number line and several sets of sample numbers. It clicked for her this time. Using the number line (sort of like using division concepts suggested by mom2bee) we saw how the GCF is essentially an interval that you can use to skip count through number alto number b. She could visualize on the line how you would have to be able to skip count through the delta (b-a) as well using the same interval. The beast workbook actually does s good job of explaining it, I think she may have just needed more examples.

:hooray: Congrats OP and daughter! So happy that she got it now!

Going back to bed now.

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M2b, the book writes it in symbols as well: GCF(a,b)=GCF(a,a-b) for integers a and b. It then asks the student to determine whether or not GCF(a,b)=GCF(a+b,a-b), giving a counterexample if it's false or a proof/explanation if it's true. Fun stuff!

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I think it's easier to imagine if you just think of factors as grouping items. If you can group a items into x groups of d (i.e. d divides a) and you can group b items into y groups of d (d being the GCD), then you can look at a-b (assuming a>b) as taking the y groups away from the x groups (since a>b implies y>x and all groups have the same size d). What you have left is obviously still in groups of d, so it's clear that d must also be a divisor of a-b.  It doesn't have to be fancy, I don't think, and might help if there is any lingering confusion (though I know you said she already got it by thinking of number lines and skip counting- sometimes an alternative way of thinking about it can just really solidify a concept, I feel! :) )

Edited by deanna1ynne
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This must be a preview of AOPS. I have heard that AOPS is a discovery approach - guiding the student to discover the principle or algorithm, rather than just teaching the algorithm.

Edited by JHLWTM
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Ok, so I had never heard of the Euclidean Algorithm. I had to google it. Up til now, my daughter had been working through Beast largely independently. I'm along for hand holding but I don't prescreen the topics or prepare any advanced introductory lessons or teaching. She just opens the books and "does the next thing." If there's a topic she gets stuck on (such as this one), I jump in and "teach."  Perhaps I should cross post or repost this under a different topic, but I wonder if that approach is sustainable for AOPS? Or will I need to be more hands on in teaching?

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FWIW, I don't have BA5. I only mentioned Euclidean Algorithm because someone else brought it up and I recognized that the topic you mentioned is a precursor to the Euclidean Algorithm. AoPS actually covers this topic in Intro to Number Theory. So I don't actually know what's in BA5, but I guess with this topic they are laying some ground work for later covering Euclidean Algorithm.

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The Euclidean Algorithm is introduced on page 44, immediately after the page in question. LOL. This stuff is absolutely intended as a thought experiment to introduce the EA.

IMO, and take this with a grain of salt because I'm using it as a supplement only, plus my kiddo who's ready for BA5 is 10 years old with great frustration tolerance...but instead of jumping in to teach, I'd jump in to learn alongside. Talk these concepts out. What I like about Beast and AOPS is the mathematical thinking. Why does this work? How? What tools do we have and how can we apply them?

I wish like anything that my children had a math circle available, but failing that, they have Beast, and the little monsters' conversations in the Guide, and my interested engagement. (My kids HATE. IT. if they think I'm doing that faux-Socratic thing where I fish for the right answer by pretending not to know it. But they appreciate it if I offer *just* as they're getting frustrated, "I think I see a thing. I have NO idea if it's the right approach, but I want to see where it goes. Mind if I share?")

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Absolutely!! By teach I really mean join in to help her think through it step by step. More than Half the time her math intuition is better than mine... I usually tell her mom has never learned x either, so let's look at it together...

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