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## 5.NBT.B.6 Transcript

This is Common Core State Standards in Mathematics. The standard is 5.NBT.B.6.

The standard reads: Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. Now, there are related standards that were started back in grade three. One of these is 3.OA.A.3, where students are supposed to use multiplication and division within 100 to solve problems. So they have had a little bit of experience with division back in third grade. Also, 3.OA.B.6, where students are supposed to understand division as an unknown-factor problem. Then back in grade 4, a related standard was, 4.NBT.B.5, where students are multiplying a whole number of up to four digits by a one-digit whole number. Notice that they are using some of the same processes, strategies based on place value and the property of operations. So, there are some similarities in the processes, and the expectation that they are supposed to illustrate and explain their thinking using either equations or some type of solid models. Also in the fourth grade, there was 4.NBT.B.6, where it’s very similar to this standard in fifth grade. The key difference was that in fourth grade they were using one-digit divisors, but now they are expected to do it with two-digit divisors. But, these are very similar standards. In the classroom, if we use the appropriate strategies and activities, we can address the standards for mathematical practices. Some of the ones that we could address through this standard are: 2. Reason abstractly and quantitatively, 3. Construct viable arguments and critique the reasoning, 5. Use appropriate tools strategically, and 8. Look for and express regularity in repeated reasoning.

Let’s focus on this idea of using relationships between multiplication and division. Now, with multiplication and division, in these contexts, really involve the same three elements: so many groups, of a certain size, and a total. And the only difference between multiplication and division is what is known and what is unknown. If we know the number of groups and the size of the groups and we don’t know the total, that’s a multiplication problem. If we know the total and the number of groups and we’re looking for the size of the group, that’s a division problem. We would use the process of division when we know the total and the size of the group and we’re looking for the number of groups. Now students need to understand the symbolism of division. Division can be expressed in several different ways. Here is one using the division symbol ÷, at this grade level; they’ll probably be using this symbolism quite often. And don’t forget the fraction bar, that’s also a symbol that can be used for division.

Let’s focus on the text in red: using strategies based on place value, the property of operations, and/or the relationship between multiplication and division. One of the adjustments that students will have to make at this grade level is dividing by a two-digit divisor. They’re not used to that, so it’s going to take a little bit of work and experience. What students might want to do is to do some scratch work on the side, because it’s very hard at this level to know how many times the divisor will go into the dividend. What did we do here? We asked ourselves how many 34’s are there in 74. And over here, we know that 34 x 2 = 68. Then, where did this come from? Well, in essence, what we’re going to do is use a property of operations; we’re using the distributive property. There is nothing that says we can’t use this for division. We are so used to using it for multiplication, but it has its place in division also. In essence, what we are trying to do is to break up the 7412 into nice divisible chunks. So 6800 is a nice divisible chunk, because it is divisible by 34. So know let’s look at 612. How many 34’s are there in 61? Two of them would be 68; that’s too much. We know that should be just a 1. But we’re dealing actually with 612, so 10 x 34 = 340, so that’s a nice divisible chunk. And now we are left with the last piece, 272. Well we can’t get any groups out of 27, so we’re going to have to look at the whole number, 272. So again, we need to do some scratch work on the side. We’re still not there, 34 x 6 = 204; that’s too small. Go a little bit further, and there it is, 34 x 8 = 272. So now we’re set. We’ve broken up the 7412 into three evenly divisible chunks that we can divide 34 into. So we have (6800 ÷ 4) + (340 ÷ 4) + (272 ÷ 34). We’ve already determined that this will be 200; 340 ÷ 4 = 10, and there are 8 34’s in 272. Notice again that we used the distributive property. It’s a great property to use here. Don’t limit the use of the distributive property just to multiplication. It does have its place with division. One of the key points here is that for whatever reason, when we were taught the standard algorithm, they gave us the idea that this 7412 was one big chunk of concrete—that’s what we have to work with, we can’t break it apart. But we can, and that’s what we did, we broke up into nice divisible chunks. And that’s really the meat and potatoes of this strategy. So our last step is simply to combine all our partial quotients to be a total of 218.

Okay, let’s look at this again. It’s a good idea for them, before they even get started, for students to write down some of the different products that come from multiplying by 34. That way, there won’t be so much guessing. With the standard algorithm, we ask ourselves how many 34’s are in 74? And over here, let’s see, it would be 68. That’ll work, so we put our 2. Then we were told to multiply it out and subtract; then the idea of bringing down the 1. Now we’re looking at 34 and 61. How many 34’s can we get out of 61? Well 68 was too big, so it’s just one. Do our multiplication, do our subtraction, and then we bring down the 2. How many 34’s can we get out of 27? We can’t, so we are going to have to look at the whole number, 272. We already know from our scratch work on the side that this is going to be 34 x 8. So, we do our multiplication, and it checks out. One of the problems with the standard algorithm is this; we did not address any of the processes here in the standard. We did not use strategies based on place value, we didn’t use the properties of operations, and we really didn’t use the relationship between multiplication and division. So the standard algorithm for division just won’t work here. We have to go way beyond that.

Let’s go back and look at this again; 34 into 74 goes twice, with some left over, but we really weren’t dividing 34 into 74, we’re dividing 34 into 7400. So we really have 200 x 34 here. Then we have the 612. What about 34 into 61? We figured it previously that that has to be a 1. But again, it’s not just 34 into 61, but it’s actually 612. So that’s really a 10, not a 1. So we subtract, and we have our 272. We already know from our scratch work that it does work out evenly, so that’s an 8. Combine all that together, and we get 218. Now notice what we did here, vertically is pretty much what happened when we use the distributive property horizontally. There’s our 6800, there’s our 340, and there’s our 272. It looks different, but the process and the thinking were really about the same. Now, if we look at what we did horizontally, with the distributive property, we can do the exact same thing with the division symbol up here. We took our number and split it up into the same divisible chunks. So we have 200 x 34 = 6800, we have our 10 x 34 = 340, and we have 8 x 34 = 272.

Let’s address the last statement of the standard: illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. Now we’ve already been using equations, but what about rectangular arrays or area models? Now depending on the resources available, it might be tough to model these really large numbers using arrays or area models. It might be a good idea to model some that are smaller, but are still four-digit. Let’s use numbers that are 2000 or less. Let’s try 1620 ÷ 36. This context could be one where we want to know how many sets of 36 there are in 1620. So here is one set of 36, but that’s going to take a while if we are going to do it one at a time. Why don’t we do 10 sets of 36 at a time? Which would be 360. So there are 10, there are 10 more sets; that puts us at 720. There are 10 more, that puts us at 1080, 10 more, and that puts us at 1440. Now, if we stop here and subtract, we still need another 180. Oh look, it’s estimation here; we know that 18 is 1/2 of 36, so instead of 10 sets of 36, it looks like I need 5. When we multiply 5 x 36, it is in fact 180. So, going back to answer the question now, we have 10 sets of 36 here, 10 sets of 36 here, and so forth, so we have 10, 20, 30, 40, 5. We have 45 sets of 36 each, so that’s our solution, 45 sets of 36. Now what about an area model? Let’s try 1334 ÷ 46, which is what we would have in this type of context. We know the area is 1334, and we know that one dimension of the rectangle is 46. What’s the other dimension? Okay, we can work with sets of 46 small square units, but that would take a while if we go one at a time. So let’s go 10 at a time. So, we have 10 sets of 46 units, we have another 10 sets of 46 units. That puts us at 920; still not enough. If we subtract, we still need another 414. And we can tell it’s not going to be 10 more because, 414 is less than 460; it’s going to be a smaller number than that. Well let’s see, let’s try 5; five 46's would be 230. But, we’re still not there; we still need another 184. If we do some more scratch work, we’ll see 4 x 46 =184. So it does come out. We’re pretty much done except we’re going back and determining what the solution needs to be. We had 10 sets of 46, another 10, 5 sets here, and 4 sets there, which gives us a total of 29. So now we know that this dimension is 29. So we’ve solved it; if we had an area of 1334, one of the dimensions is 46, and then the unknown dimension had to 29.

With this standard, again the focus is on using strategies based on place value and the properties of operations, and students are expected to illustrate and explain their calculations, either through equations, rectangular arrays, or area models. So again, this standard takes students way beyond the standard algorithm for division.