Dividing 3-digit by 2-digit numbers using partial quotients
At about 200-203s the narrator says '15 times 10 is 100' (should be 150). A non-pausing note flags the slip.
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The video finds 497 / 15 using the partial quotient method. It lists easy multiples of 15, takes away 15 x 30 = 450 and then 15 x 3 = 45, and adds 30 + 3 to get 33 remainder 2. It then solves it again with different chunks (20, 10 and 3) to show you get the same answer.
You'll be able to
- divide a 3-digit number by a 2-digit number using partial quotients.
- use multiples like 15 x 3 = 45 to find 15 x 30 = 450.
- show that different choices of chunks give the same quotient and remainder.
Key words
- partial quotients
- A way to divide by taking away easy groups of the divisor and adding up how many groups you took.
- multiple
- The result of multiplying a number by a whole number, like 15, 30 and 45 for 15.
- estimate
- A sensible guess of how many times one number fits into another.
- remainder
- What is left over when no more groups of the divisor fit.
Transcript
- [Lecturer] All right, what we're gonna do in this video is we're going to try to figure out what 497 divided by 15 is. And there's many ways that we can approach this, but we are going to use the Partial Quotient method. Now that sounds super fancy, but really all it is is trying to make some estimates of how many times 15 can go into this number without going over it and then figuring out what's left over and then figuring out how many times 15 can go into what's left over, and then we'll total up all of the numbers of times that 15 goes into it.
So what do I mean by that? Well, before I even go into it, let's just think about some multiples of 15. So, obviously, we know 15 times 1 is equal to 15. 15 times 2 is equal to 30. 15 times 3 is equal to 45. And this is useful. I could keep going. I could go to 15 times 4, et cetera.
But just knowing this, I could also say, well, 15 times 10 is 150, 15 times 20 is 300. 15 times 30 would be 450. So if we wanted to use Partial Quotients over here, one way is we could go very quickly and get as much of this 497 as possible. I just said that 15 times 30 is 450.
15 times 30 is equal to 450. So we could do that. We could say, all right, if we were to say, and the way the notation I will use right over here, I will say, okay, I know that 15 times 30 is equal to 450. So if I go in 30 times, I'll write the 30 right over here, then I'm going to have 450.
And then what am I going to have left over? Well, 497 minus 450 is going to be, we get a 7 here, we get a 4 here, it's going to be 47. So I still need to go, I still have more 15s that I can fit into this 'cause 15 definitely goes into 47. Well, we know that 15 times 3 is 45, so why don't we try to go, lemme write this over here, 15 times 3 is equal to 45.
And so let's go in exactly three times. Actually, let me write this right down here. 15 times 3 is equal to 45. So if we were to put three more 15s in there, we're going to get 45. And then what do we have left over? Well, 47 minus 45 is equal to 2, and we definitely can't fit any 15s into 2.
So that's just going to be our remainder. So how many times is 15 going to 497? Well, first, we went 30 times, and then we went three more times. So it goes a total of 30 plus 3 times, or it goes 33 times. We could write it, like, we could write 33 up here, or we could just write that 497 divided by 15 is equal to 33 with a remainder of 2.
Now I wanna be clear that this isn't the only way to do it. This is what's neat about the Partial Quotients method. You just make estimates don't go over 497 and you'll still get to where you need to go to. For example, I'll just do another quick example. Maybe I'll do it right over here and I'll do it a little bit smaller so I don't run outta space.
So 15 into 497. Let's say I wanna be a little bit more careful and I say, all right, well, I know that 15 times 10 is 100 and, well, lemme start with 15 times 2. 15 times, or 15 times 20 is equal to 300. So let me go 20 times and then that will get me to 300. If I subtract here, I get 197.
And then I say, okay, well, I know that 15 times 10 is 150. So now if I subtract, I get the 47, and then this is where I could say, okay, 15 times 3 is 45, and if I subtract, I get a remainder of 2 again. And so I have, how many times did I get the 15 into the original 497?
I had 20 times and then another 10 times, and then another three times. If I add these 20 plus 10 plus 3 is, once again, this 33, and once again, I have a remainder of 2. So many different ways that you can approach this with Partial Quotients.
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