This is Common Core State Standard support video for mathematics, Grade 4. This one is in the domain that focuses on number and operations in base 10, and the particular standard is point six. So, this standard states: find whole number quotients and remainders with up to four-digit dividends and one-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 models.
Now, the typical context for whole number division is one of these two situations. We know the total and the number of groups, and we’re looking for the size of the group. Or we know the total and the size of the group, and we want to know how many groups could we make from that total of that certain size. So, let’s start off with an example so we can make sense of the algorithm; and let’s say that we have a total of 96, and we know that we want seven groups. What would the size of the group need to be?
So, we’re going to take our two-digit dividend and divide it by a one-digit divisor of seven. So, if we go through, and we follow the typical steps of the standard algorithm for long division; let’s see, 7 into 9, let’s see, that will be one time; subtract, that’s 26, 7 into 26 would be 3 with 5 left over. So, we have a remainder of 5. So, our answer would be 13 remainder 5 or 13 and five sevenths, depending on the form that you want the students to put the quotient in.
One of the shortcomings about the standard algorithm is that we sort of lose the idea of place value, and that is one of the strategies that we want to use, to focus on place value. Now, what really happened here with that initial 1, we really weren’t multiplying 1 times 7. That was really 10 times 7. So, this was actually 70. To get a better handle and a better picture of that, let’s rewrite this to where 96 is 90 plus 6. Well, let’s see, how many groups of seven can I get out of 90? Let’s see, that would be 10. Okay, so that would be 70, and then we have 20 plus the 6 that we have over here is the 26 that we had there. And so, now, let’s see, we can get three groups, let me bring this down; we can get three groups of seven out of that with five left over. So, again, looking at it from this perspective here gives us a little bit better handle of what’s really happening with respect to the place value.
It’s always a good idea to include some solid concrete objects to see what’s really happening with the operation. So, written down here in equation form, this is what we did just a minute ago. We took the 96, we broke it up to 90 plus 6. Then, using our distributive property, we’re actually taking the 90 and dividing by 7, and the 6 and dividing by 7. So, we actually have two division operations. So, we look at the first part of the problem. If we look at 90 divided by 7, we’re looking at, well, how many groups of seven can we get out of 90? And if we take the solid manipulatives, and do this to where we have groups of seven, well, we can get a total of 12 whole groups of seven out of this.
But, we have a problem because we have some left over over here. We don’t have enough to make another group of seven. We’re missing one right here. But, we can take care of that. We can, we noticed that we have some over here to work with. So, we can steal one of those and bring it over here. So, now, that gives us an additional group of seven. So, instead of 12 groups of seven, now we have 13 groups of seven. And over here, well, we don’t have a whole group of seven here. We’re missing two. So, we don’t have a whole group of seven. We have five out of seven, or five sevenths of a group of seven, because again, we didn’t have enough. So, our solution then, would be 13 with a remainder of 5 or if you wish for students to put the answer, the quotient, differently, we would have 13 whole groups of seven plus five sevenths of another group of seven.
The distributive property is one of the most powerful properties that we have to work with in mathematics. One of the things that we’re not used to is to use the distribute property with division. But, there’s a lot of power in using this with division, not just multiplication. The key here is to use sensible grouping where we’re not bound to place value. How can we take the 96 and break it up into smaller chunks that are divisible by seven, where you can take a nice whole number of groups of seven out of it?
So, if we take the 96, well, let’s see, that’s 70 plus 26, because again, the 70 is easily divisible by seven. Okay, once we’ve done that, now, what about the 26? Oh, well, I can break that down to 21 plus 5. So, now, using our distributive property, we are taking the 96, and we broke it down to three component parts. So, now we actually have three division problems. We have 70 divided by 7, plus 21 divided by 7, plus 5 divided by 7.
And if we give the students a little bit of help and use some manipulatives, well, let’s see, we’re working with groups of seven. How many groups of seven? Well, here we have 10 groups of seven. And then, let’s see, over here we can get three groups of seven out of that, and here they are. And then we don’t have enough to make another whole group. We’re short two, so we only have five sevenths of another one. So, we reach the same solution that we did earlier, but slightly differently where we, again, used grouping based on the divisibility of seven rather than place value. So, again, we ended up with 10 plus 3, 13, and five sevenths as before.
Let’s ratchet this up a little bit and go to a three-digit divisor. So, let’s say the situation is, we have 872 for our total. We want the groups to be size 6 and we want to know how many groups we need for that. Well, let’s forgo doing the standard algorithm, because that’s something that teachers can do in the classrooms separately. But let’s supplement that to where, again, the place value is really understood. So, this is really 800 plus 70 plus 2. Well, let’s see, how many sixes can I get out of 800? Well, I’m dealing with hundreds here, so, well, let’s see, I can get 100 groups of six, which would be 600; and I have 200 left over, plus the 70 over here, which gives me 270. Okay, let’s see. Well, let’s see, 40 groups of six would be 240. So, we have 40 sets of six. That’s 240. And now, we have 30 left over, plus the 2 that we still have here. That gives us 32. And, let’s see, we can get five groups of six out of the 30, well, out of the 32. And we have a remainder of 2. So, our solution then, if we combine all of these together, would be 145 with a remainder of 2, or written as a mixed number, two sixths, which of course, would simplify to one third.
Let’s try this again, but let’s use the distributive property. Again, we don’t want the students to fall into this mistaken notion that the 872 is one big solid chunk of concrete that we cannot break up into smaller chunks. We can. So, we’re dividing by six. We want groups of six. Well, let’s see, well, we can break that down to 600, because 600 is easily divisible by six. So, we’ve done that. Now, we need to worry about the 272. That breaks down to 240 plus 32. And now, we need to break down the 32. And that breaks down to 30 plus 2. Notice from the previous work that we just did, you can kind of see the tie-in between the place value that we used here and the place value here, and the two ideas, the two approaches, start sort of meshing together.
So, now, using our distribute property, we take this one division by six and split it up into separate operations. So, we have 600 divided by 6 (How many sixes can I get out of 600?), plus 240 divided by 6, plus 30 divided by 6, plus 2 divided by 6. Well, let’s see, we have 100 here, 100 groups of six. Here we can get 40 groups of six. Here we can get five more groups of six. And then, over here, we can’t take a whole number of sets out of this. So, we just have part of a set of six here. So, again we get 140 plus 5, 145, plus 2 left over out of a group of six. And, again, you can have the kids simplify this to 145 and one third. Or, again, if the students are at the stage where you’re dealing with just remainders, you can get 145 groups of six with a remainder of 2.
Let’s try one more example. The standard calls for up to a four-digit dividend. So, let’s try 2531 as our total. We want nine groups, so we need to find the size of the group. Again, let’s forgo the typical standard algorithm, and let’s do this based on place value. So, we’re going to divide 9 into 2000 plus 500 plus 30 plus 1. Well, let’s see, going by the first place value, that’s 2000. Let’s see, I can get 9, that’s 2 times 9 is 18, so let’s try a 2, but this is really 200. That gives me 1800. That’s 200 plus the 500 that I have here. So, that’s a total of 700. So, now we’re dealing with how many groups of nine in 700? Well, let me see. That would be seven, eight; yes, 7 times 9 is 63.
So, let’s just try, but that’s not a 7, that’s really 70. So that gives us 630, and we have a remainder of 70 here. Okay, here it gets a little interesting, because 70 plus 30 is 100. Let’s see, 9 into 100, how many nines in 100? We can get 11, 11 times 9 is 99; and 1 left over, but don’t forget the 1 that we still have here. So, that’s a 2. Can’t get a whole group of nine out of a two, so that’s it. So, we have 270 plus 11, that’s 281. And we had a remainder of 2, so, we have two ninths of a group of nine. So, we have 281 and two ninths.
Notice that this one got a little messy. It got a little bit complicated over here, because I ended up with 100 for this total of 70 and 30. So, here, place value almost made the problem a little bit more confusing for the kids. So, again, sometimes it’s better to go with some type of sensible grouping that you’re not bound to place value, again, using the power of the distributive property. So, if we take 2531 and we break it down, well, again, 9 times 2 is 18. So, a good way to break down the 2531 is 1800 plus 731. Now let’s worry about the 731. Let’s see, 9 times 8 is 72. So, we can break this down to 720 plus 11. Then, we’re worried about the 9, sorry, the 11. That breaks down to 9 plus 2. Again, if you look back at the work that we just did, notice the similarities. I’m starting to recognize some numbers, the 1800, the 720, and the 11.
Alright, this would be very tough to do with manipulatives, but if you have the manipulatives to handle this size of a number, where you have some blocks of 1000 and 100 and so forth, then please make sure that you connect what we’re doing here symbolically with a solid concrete representation of this, similar to what we did with the first example. So, again we’re using the power of the distributive property. We broke down the 2531 to smaller chunks that are nice because they’re divisible by nine. So, we start off with our first one. Let’s see, that’s 200 sets of 9 here, plus, let’s see, 8 times 9 is 72, so this is 80 more sets of 9. Nine divided by 9, well, that’s just 1 group of 9. So, there’s 281 so far.
And then over here, we don’t have enough to make another group of nine. We have just part of one. So, that’s two ninths of another set of nine, or just remainder 2, again, depending on where your kids are. So, again, this is just an example that shows one more time the power of the distributive property, that one alternative to your standard algorithm is to use the distributive property, and use sensible grouping where you break it down into chunks where the numbers that you are dealing with are divisible by that one-digit divisor.