# Lesson Video: Multiplying Two-Digit Numbers: The Column Method Mathematics • 4th Grade

In this lesson, we will learn how to use the standard algorithm to multiply a two-digit number by another two-digit number and regroup when necessary.

18:29

### Video Transcript

Multiplying Two-Digit Numbers: The Column Method

In this video, we’re going to learn how to use the standard written method or algorithm to multiply a two-digit number by another two-digit number. And where we have to, we’re going to regroup to help us find the answer.

Let’s imagine that we want to multiply together 35 and 24. Now we know that we can partition 35 into 30 and five. And if we wanted to multiply 35 by a one-digit number like four, for example, we would just need to make sure that both parts were multiplied by four: five times four and then 30 times four. Now, with the question that we’ve got here, we’re not multiplying 35 by a one-digit number. We need to multiply it by a two-digit number. And just like we can think of 35 as 30 and five, we can partition the number that we’re multiplying by 24 into 20 and four. So as well as making sure that we multiply five by four and 30 by four, we need to make sure that we also multiply both parts of 35 by 20 too.

In other words, we need to find out what five times 20 is and also what 30 times 20 is. Now, this diagram isn’t how we’d find the answer. We’ve just drawn it here because it’s a useful way to show us what we need to do. To find the answer, we need to multiply every part of the number 35 by every part of the number 24. Hopefully, you can see by our four arrows that there are four multiplications we need to do. So that’s exactly what we’re going to do now as we use the column method. It’s going to involve working out the answers to four separate multiplications and then adding up the products at the end to find the overall total. And because this is the column method, the very first thing we can do is to write our calculation out in columns.

Just for our first example, let’s use these colored column headings, and also we’ll use different colors for our digits. So 35 is three 10s and five ones. And we want to multiply this by 24 which is two 10s and four ones. Now just as we said at the start, we need to multiply both parts of 35 with both parts of 24. So we’re going to have to do four multiplications. And to begin with, let’s multiply everything by the four ones in 24. So, first of all, what’s five times four? My straightforward one to begin with: five times four is 20. And we can write the number 20 then underneath the equals sign as the first of our four multiplications. And just so we remember what we’ve done, we can make a note of the multiplication we’ve done by the side, which was five times four.

Now, can you see what else we need to multiply by the four part of 24? We need to multiply this digit here by four too. Now it would be very easy to look at this and to think to ourselves what’s three times four because you can see the digit three. We need to multiply it by four. But we need to remember that this digit three is in the tens place. We need to find the answer to 30 times four. Now, we can actually use the idea of three fours to help us, though. We know that three fours are 12. So three 10s multiplied by four will be the same as 12 10s, which is 120. Now, if we pause there for a moment, can you see what we’ve done already? We’ve multiplied everything in the number 35, that’s the five and the 30, by the digit four in 24.

Next, we need to multiply everything by the tens part of 24. And we start with five times 20. Now again, if we look at these digits, it might look like we’re multiplying five by two. But we can see that that our digit two is in the tens place. This is five times two 10s. But once again, we can use facts we already know to help us. Five times two is 10, so five lots of two 10s must be 10 10s, which is the same as 100. Five times 20 is 100. So after multiplying five by 20, the very last thing we need to do is to multiply 30 by 20.

Now, out of the four different multiplications that we do when we use column multiplication, this last one is probably the trickiest. This isn’t because they’re particularly big numbers or anything like that. It’s just because it’s probably the easiest one to make a mistake with. We could just look at the three and the two and think three multiplied by two is six, but we need to remember that both these digits are in the tens place. This is 30 multiplied by 20. But we could still use the fact three times two is six to help us here. If three times two is six, then three 10s multiplied by two will be the same as six 10s or 60, and three 10s multiplied by two 10s will be 10 times greater again. It’ll be the same as 60 10s or 600.

Can you see the pattern there in our questions and answers? So 30 multiplied by 20 equals 600. So we’ve multiplied both parts of 35 by the four in 24. And then we multiplied both parts of 35 by the 20 in 24. Now all we have to do is to find out what the total product is. In our ones column, we’ve only got these zeros, so we can put a zero in the ones place. If we add the tens, we’ve got two and two and then two zeros, so that’s four 10s altogether. And if we add the hundreds, we’ve got one plus one plus six. So that’s eight 100s. And so, we can say the product of 35 and 24 is 840. Let’s have a go at practicing this column method now with some questions.

Calculate the following: 29 multiplied by 64 equals what.

In this question, we’re given a pair of two-digit numbers to multiply together, and we’re given a really big clue as to how to do this because the way that this calculation is being set out is by writing both numbers on top of each other. This means that digits are in columns. One way of describing this is as the column method. And by writing the numbers like this, it helps us to split them up into their tens and ones. Both parts of the number 29 need to be multiplied by both parts in the number 64. Let’s make a plan of all the multiplications we’re going to need to do.

So to begin with, we’re going to need to multiply every part of the number 29 by the four ones in 64, so that’s nine times four and then 20 times four. That way, we’ve multiplied 29 by four altogether, haven’t we? Then we need to multiply every part of the number 29 by the six 10s in 64. So we’ll start off by working out nine times 60 and then, finally, 20 times 60. Then we’ll have also multiplied all the parts of 29 by 60. And if we then can combine all our parts together, we can find the answer. So to begin with, let’s multiply by our four ones. Now we know multiplying by four is the same as doubling and then doubling again. So finding out nine times four is the same as doubling nine to get 18 and then doubling 18. 18 doubled is 36, so we know nine times four is 36.

Next, we need to multiply 20 by four. Remember, this digit two doesn’t have a value of two. It’s in the tens place. It’s worth 20. Again, we can use doubling to help us. 20 doubled is 40, and then 40 doubled is 80. So we know that 20 times four is 80. Now we need to multiply everything by the six 10s in 64. So what’s nine times 60? Well, we can use our knowledge of place value to help us here. Nine sixes are 54. So nine lots of six 10s equals 54 10s, which is the same as 540. Finally then, we need to multiply 20 by 60. Now, to help us, we can recall what two 60s are, and that’s 120. So by trying to find out what 20 lots of 60 are, our first factor has increased 10 times. Instead of two times 60, we’re actually looking for 20 times 60.

Our answer then is going to be 10 times larger. And we know that to find a number that’s 10 times larger than another one, we just shift the digit one place to the left, so 120 becomes 1200. So now we’ve multiplied each part of the number 29 by the ones and then the tens of the number 64. To find our overall answer then, we just need to add these partial products together. So to start with, if we add our ones, we can see the digit is six and all the other digits are zero. So we’ve got six ones. In the tens column, we got a few more to think about. We’ve got three 10s plus eight 10s plus another four 10s.

Now, how would you add these quickly? Perhaps we could put the three and the four together to make seven 10s. And we know that eight 10s plus eight 10s will be 16 10s. So if we add eight 10s to seven 10s, it’s going to be 15 10s. So we’re going to need to exchange 10 of our 15 10s for 100 and then five 10s. Adding the hundreds then, we’ve got five 100s plus another two 100s. That’s seven 100s. Not forgetting the one that we’ve just exchanged, that’s eight 100s altogether. And there’s just the 1000 in our thousands column. So in this question, we’ve multiplied together a pair of two-digit numbers using the column method. This helps us to make sure that every part of the number 29 is multiplied by every part of the number 64. 29 multiplied by 64 equals 1,856.

What number can replace the question mark in this calculation? Complete the calculation to solve it.

The calculation that’s mentioned in the question is this one here. It looks like the column method has been used to multiply a pair of two-digit numbers. 33 multiplied by 30. Oh! We’ve got a digit missing here. There’s a question mark. And the first part of this problem asks us what number can replace this question mark. Now we could look at this calculation and say to ourselves, “There are lots of possible answers. The missing digit could be anything from zero up to nine.” But you know, this isn’t true because we’re given one more piece of information. We can see that someone has already started working out the answer to this multiplication, and they’ve already found the partial product 132.

Now, when we’re using the column method like this, usually the first thing that we do is multiply everything by the ones in the second number. So to start with, we’d multiply the three in 33 by the ones in the second number; then we’d multiply the 30 in 33 by those ones in the second number. Then we do exactly the same, this time multiplying by the tens in the second number, so that would be four multiplications altogether. But can you see the way that this calculation is being set out? There’s only space for two partial products. In other words, the person that’s working out the answer is going to multiply 33 all in one go. So that’s 33 multiplied by the ones, which, of course, we don’t know at the moment, and then 33 multiplied by the tens. So that’s 33 times 30.

Now that we know what’s happening in this working out, we can use it to find out our missing number: 33 times what gives us an answer of 132. Now, the important digit we need to think about here is the digit two. What digit could we multiply our three ones by that would give us an answer that ends in two? Well, obviously, two is less than three. It’s not a multiple of three, so we need to think about a two-digit number that ends in a two. And we know that three times four is 12, and 12 ends in a two. Let’s see whether 33 times four is correct. As we’ve said, three times four is 12. That’s the same as one 10 and two ones. And because three times four is 12, we know three 10s times four must be 12 10s. We’ve got one 10 as well underneath we need to remember to include, so that’s 13 10s. And there’s our number 132.

The calculation is clearly 33 times 34, and our missing digit is four. Finally then, we’re just asked to complete the calculation to solve it. What is 33 times 34? Well, we’ve worked out the first partial product, so now we just need to work out the second. We need to multiply 33 by the tens digit in 34. In other words, 33 times 30. Now we know this number we’re multiplying by 30 is only 10 lots of three. So why don’t we multiply 33 by three and then use this to help? Three times three is nine, and three 10s times three is nine 10s or 90. So if 33 times three is 99, then 33 times three 10s will be the same as 99 10s, which is 990.

So we’ve multiplied 33 by four. Then we’ve multiplied 33 by 30. Now we just need to add these two partial products together. Two ones plus zero ones is two ones. Three 10s plus nine 10s equals 12 10s, which is the same as 100 and two ones. Then 100 plus nine 100s is 10 100s plus the one that we’ve exchanged equals 11 100s, which is the same as 1,100. And we can just write that 1000 directly into the thousands place. This was an interesting question because as well as using the column method, we had to use what we knew about it to help find a missing digit. The number that replaces the question mark in the calculation is four, and 33 times 34 is 1,122.

Those questions gave us some really good practice there. But before we finish this video, let’s look at one mistake that we really need to avoid. It’s quite easy to make as well. Here, we can see that this girl’s trying to multiply 26 by 14, and she’s ended with a total of 40. But it just doesn’t look right to her. 26 times 14 is 40? She’s made a mistake here, and it’s a very easy mistake to make. Can you spot what it is? If we start at the end and work backwards, we can see that she has actually added those digits correctly. Four plus eight plus six plus two equals 20. And then two 10s plus the two 10s that she’s exchanged is 40, so she’s done this addition part correctly.

The problem must be with the multiplication at the start. The first partial product she’s got is 24. And we can see where she gets this from. She’s begun by multiplying by the ones, and six times four ones is 24. Next, she’s multiplying this digit two by four. And can you see what she’s done? She’s given this digit a value of two. Two times four is eight, but we know this digit two isn’t worth two. It’s worth 20. It’s in the tens place.

So she should have been multiplying 20 by four, which gives the answer 80. Now it’s a really easy mistake to make, and this girl has carried on making all the way through. But it’s something we need to avoid. Instead of six times one, she should have calculated six times 10. And instead of two times one, she should have found the answer to 20 times 10, which gives us a very different result. Always be careful with your place value.

So what have we learned in this video? We’ve learned how to use the standard written method to multiply a pair of two-digit numbers together.