Hi! Welcome to Math Antics. In our video called “Long Division” we learned how to do division problems that had long multi-digit dividends. They key was to break up a big division problem into a series of smaller and easier division steps. And that involved trying to divide the dividend one digit at a time… digit-by-digit. And in the examples we saw, going digit-by-digit was pretty easy because we only had one-digit divisors. But, what if you need to use that division method for problems that have bigger divisors? …like if you’re dividing by a two or three-digit number? In this lesson, we’re gonna learn how you handle problems like that. The good news is that you kind of already know what to do, you just may not realize it yet. To see what I mean, have a look at these two division problems. They both have the same dividend and both have a one-digit divisor. But these divisors are different numbers, and as you’ll see, that’s going to effect our digit-by-digit division process. To solve this first problem, we start by asking, “How many ‘2’s does it take to make 5 or almost 5?” Or you can think of it as, “How many ‘2’s will fit into 5?” And it’s easy to see that the answer is 2. So we put a 2 as the first digit of our answer. Then we multiply 2 times 2 which is 4, and we subtract that 4 from the 5 which leaves us a remainder of 1. Now we move to our next digit and we need to bring down a copy of it to combine with the remainder from the first digit. Then we ask, “How many ‘2’s will make 12?” That’s easy: 6. So we put 6 as the next digit of our answer. 2 × 6 = 12, and 12 minus 12 leaves no remainder. And finally, for our last digit, even though there was no remainder, we can bring a copy down and ask, “How many ‘2’s will make 8?” And the answer is exactly 4. 4 × 2 = 8 which again leaves no remainder. There, we went digit-by-digit and broke our problem up into three division steps (one for each digit) and we got our answer: 264. Now let’s solve the next example. And right at the start, you’ll see we have a bit of a problem. When we ask, “How many ‘8’s does it take to make 5 or almost 5?” The answer is NONE! And that’s because the first digit (taken by itself) is LESS than the divisor. 8 is TOO BIG to divide into 5. So what do we do? Well, Instead of just trying to divide the first digit all by itself, let’s group the first two digits together? If we group the 5 and the 2 together, then our first step will to ask, “How many ‘8’s will make 52?” That’s better… 8 will divide into 52 about 6 times. So we’ll put a 6 in our answer line, right above the 2. Why does it go there? Because we had to skip the first digit and group it with the 2. If we wanted to, we could have put a zero above that first digit since the 8 wouldn’t divide into it ANY times, and if that helps you keep track of which answer digit you’re on, then that’s a good idea, but it’s not required. So, 6 × 8 = 48 and then 52 minus 48 gives us a remainder of 4. Now, we only have one digit left to divide so we bring down a copy of it to combine with the remainder and ask, “How many ‘8’s will make 48?” We know the answer to that is 6 also. 6 × 8 is 48 which leaves no remainder. There… our answer is 66. Did you notice the difference between these two problems? We wanted to go digit-by-digit in both problems, but in the second problem, the divisor was bigger than the first digit of the dividend, so we had to start out by going two-digits at a time in that case. And that helps us see something really important about this traditional long division method. You don’t always HAVE to go one digit at a time. You can break the dividend up into bigger ‘chunks’ of digits if you want and apply the same procedure to those bigger chunks. You could go two or three digits at a time or even try to divide the entire dividend all in one step! And taking bigger chunks of the divided usually results in fewer division steps. Notice that there were three steps in the first problem but only two steps in the second problem. Fewer steps!? I like the sound of that! That seems like a lot less work!! Ah yes… fewer division steps does SOUND better, but it’s really not. That’s because the more digits you group together, the harder that division step will be. I thought it sounded too good to be true. It’s kind of like climbing stairs. When you have a lot of small steps, each one is easy to climb. But with only a few big steps, each one can be a challenge of its own. That’s why we always TRY to go just one digit at a time. If you only have to divide into one or two digits of the dividend at a time, it’s much easier because all the answers to those smaller division steps can be found on the multiplication table, (which you have memorized, right?) But when we have to go 3 or 4 digits at a time, it’s a lot harder to figure out the answer of each step. Okay, but how does that relate to two-digit divisors? Ah… as you’ll see, two-digit divisors force us to take bigger steps! To see what I mean, let’s try solving two new division problems that have the same dividend as before but two new divisors, and both of these are TWO-digit divisors. In this first problem, we COULD start by asking, “How many ’24’s will fit into 5?”, but since our divisor now has two digits, we already know that NO one-digit chunk of the dividend will be big enough for that to divide into. So, because we have a two-digit divisor, we automatically need to group the first two digits and ask, “How many ’24’s will make 52?” This is trickier because multiples of 24 are not on our multiplication table. Instead, we have to figure it out by estimating (or good guessing). Because we know that 2 times 25 would be 50, 2 is a really good estimate for the first digit of our answer. 2 times 24 is 48 and then when we subtract 48 from 52, we get a remainder of 4. Okay, so far so good… we’ve already dealt with the first two digits of the dividend, so now we bring down the last digit to join the remainder and ask, “How many ’24’s will make 48?” That’s easy, it’s 2 again because we just saw that 2 times 24 is 48, so that will leave no remainder. So the answer to this first two-digit divisor problem is 22. Now let’s have a look at the next problem. It’s also got a two-digit divisor, so we’ll start the same way. We’ll start with a two-digit chunk of our dividend and ask, “How many ’88’s will it take to make 52 or almost 52?” Uh Oh! …see the problem? Even though both are two digits, this won’t work because 88 is already greater than 52. And that means, we’re going to have to take an even BIGGER chunk of this dividend. We need to group the first THREE digits together! But… that’s just like doing the whole problem at once, without breaking it into any steps! Yep… and that’s why division problems with big divisors can get difficult. When you have a two or three-digit divisor, each step might be as big as the whole long division problem and it can take a lot of trial and error to figure out. In fact, if we had our way here at Math Antics, when division problems get that complicated, we’d just let students use calculators to solve them. What do we want? Calculators! When do we want them? Whenever we have long division with two or more digit divisors! Okay, but what if we don’t get our way and you need to solve this problem without a calculator? What’s the best strategy? Well, a little estimating will help us make much better guesses at our answer. The numbers 88 and 528 are kind of hard to work with, but if we made estimates of those numbers… Like if we change them to 90 and 500, that would make it easier to estimate the answer. Since 100 would divide into 500 exactly 5 times, that means that 90 will divide into 500 at least that many times. So, let’s make 5 our first estimate for the answer. To check to see how good that estimate is, we multiply 5 by 88 and then subtract that from 528 to see what the remainder is. Now 5 × 88 is kinda tricky on its own, so you may want to use scratch paper to work it out. 5 times 88 is 440 and when we subtract 440 from 528, we get a remainder of 88. Hmmm… it looks like our estimate was too low. Whenever the remainder is greater than or equal to the divisor, it means we underestimated the answer. In fact, since our remainder is EQUAL to the divisor, it means we could have divided exactly one more 88 into 528. So we should have picked 6. And if you multiply 6 times 88, you’ll see that it’s 528. So as you can see, even though the division procedure is basically the same in all these cases, the value of the divisor makes a big difference on our division steps. Whenever the divisor is bigger than the part of the dividend that we’re trying to divide, it means that we need to group more digits and take bigger division steps. Let’s try one more much longer two-digit divisor problem. 817,152 divided by 38. I’m gonna work through this kinda fast, so you may want to re-watch it a couple times if you have trouble following it. Since we have a two-digit divisor, we start with the first 2 digits of the dividend and ask, “How many ’38’s will it take to make 81?” Again we’re gonna use rounding to help us estimate the answer. 38 is close to 40, and 81 is really close to 80. And since 80 is 2 times 40, my estimate for the first answer digit will be 2. 2 × 38 = 76. And 81 minus 76 leaves a remainder of 5. We know our estimate was just right because 5 is less than our divisor of 38. Now we move on to the next digit. We bring a copy of it down and combine it with our 5 and ask, “How many '38's will it take to make 57?” That one is easier to estimate. …just 1, because it’s easy to see that two ’38’s would be too big. 1 × 38 = 38. And 57 minus 38 leaves a remainder of 19. On to the next digit… we bring down a copy of the 1 and now we ask, “How many ’38’s will it take to make 191?” That’s a bit tougher. To estimate, I’ll round those numbers to 40 and 200. And I know that five ’40’s makes 200, so 5 is my estimate for the next answer digit. 5 × 38 = 190. And 191 minus 190 leaves a remainder of 1. Moving on... we bring down a copy of our next digit and ask, “How many ’38’s will it take to make 15?” Uh oh. 15 isn’t big enough to be divided by 38. But don’t worry, we already know what to do when this happens. Whenever we’re trying to divide a bigger number into a smaller number, we just put a zero in the answer line and move on to the next digit. We bring down a copy of the 2 and combine it with our remainder of 15. Now we ask, “How many ’38’s will it take to make 152?” To estimate this one, I’m going to round those numbers to 40 and 160. And since 4 × 40 = 160, I’ll put 4 in the answer line as my estimate. 4 × 38 = 152. And 152 minus 152 leaves no remainder. And we’re done! Wow! That was a lot of work! But did you see how much rounding helped us out? We made good estimates each time by rounding the numbers we were working with. Alright, now you know that the long division procedure works the same for two-digit divisors. It’s just that each division step will involve two or three digits of the dividend. And, since each of those bigger steps is harder to figure out, you’ll want to use estimating to help find the answers. And while it’s good to know how to do complex division problems like this, we still think that complex division problems are a job for your calculator. So try a few practice problems, but don’t wear yourself out doing really long division like this. After all, the reason we study math is to become good problem solvers and to be able to understand all sorts of important math ideas, and there’s a lot more to math than division! As always, thanks for watching Math Antics and I’ll see ya next time. Learn more at www.mathantics.com
