Question Video: Understanding Place Value When Performing Long Division with a Three-Digit Divisor

Video Transcript

We want to divide 6025 by 241 using the standard division algorithm. This question has six parts. And we’ll take each part in turn, beginning with “what result do you expect?” a) Around two, b) around three, c) around 30, d) around 200, or e) around 3000.

Our first step requires us to do some estimating. We can round 6025 down to 6000 and we can round 241 down to 200. We do this because we’re rounding to the hundreds place. To the right of the hundreds place, the tens place, there’s a four. This is less than five, which means we would round this number down. When we were dealing with 6025, we rounded to the nearest 1000. And we looked to the right of the thousands place to the hundreds, where there was a zero, indicating that we should round down.

And now, we need to consider dividing 6000 by 200. We know that 200 is 100 times two. We could divide 6000 by 100 which is 60. We then take that 60 and divide it by two, which gives us 30. And that tells us we can expect a result for 6025 divided by 241 of around 30. Now that we have a good estimate, we can move on to the second part.

The diagram below shows the first stage of the division algorithm. The digit 𝑎, between one and nine, represents how many 241s there are in 602 tens. What is the place value of 𝑎 in the quotient?

First, let’s think about the place value. Underneath the division bar, we have 6025. The digit six is in the thousands place. There are zero hundreds, two tens, and five ones. And our divisor 241 has a two in the hundreds place, a four in the tens place, and a one in the ones place. Now, when we use the standard algorithm, we line up the quotient with its place value that’s underneath it. In our problem, the dividend 6025 has an 𝑎 above the tens place. Since our question asked what is the place value of 𝑎 in the quotient, we can say that 𝑎 is in the tens place.

The next part of our question wants to know what is the value of 𝑎.

We’ve already said that 𝑎 is in the tens place. This 𝑎 is answering the question, how many times does 241 go in to 602 tens? It can be helpful to remember that this means we’re looking for a value of 𝑎 that multiplies by 241 and gets us as close to 602 as possible without going over that amount. We could try three because 200 times three equals 600 and that seems pretty close. But remember, we also have four tens and 40 times three is 120. And, of course, one times three is three.

When we add up these three values, we get 723. It also means that 241 times three is greater than 602. And therefore, 𝑎 cannot be three. And so, we need to try again with the smaller 𝑎 value. We can try two. 200 times two equals 400. 40 times two equals 80 and one times two equals two. When we add those together, we get 482. And we can say that 241 times two is a less than 602. Since there are no whole numbers between two and three, the value of 𝑎 must be two.

Since we found that there are two 241s in 602, 𝑎 equals two, or 241 times two tens in 602 tens. The number 𝑏 is two times 241 tens. Which of the following is true? a) 𝑏 equals two times 241, b) 𝑏 equals two times 241 plus 10, c) 𝑏 equals two times 241 minus 10, d) 𝑏 equals two times 241 times 10, or e) 𝑏 equals two times 241 times 100.

Let’s go back to the beginning when we were thinking about place value underneath this division bar. We have thousands, hundreds, tens, ones. And we said that the 𝑎 value, the value above the tens place, was tens. After which, we found out that 𝑎 equals two. In the same way that order matters above the division bar, it also matters underneath. The value 𝑏 is in the tens place. We find 𝑏 by multiplying two times 241. We did that on the previous slide, and we got 482. But we know that this is 482 tens. 482 tens equals 482 times 10. And we know that we could find 482 by multiplying two times 241. And so, we can say that 𝑏 equals two times 241 times 10, which is option d in the answer choices.

The number 𝑐 is the remainder of the previous division. Which of the following is true? a) 6025 equals two times 241 plus 𝑐, b) 6025 equals two times 241 minus 𝑐, c) 6025 equals 20 times 241 plus 𝑐, d) 6025 equals 20 times 241 minus 𝑐, or e) 6025 equals 200 times 241 plus 𝑐.

We know that 𝑐 is the remainder from the previous division. So let’s focus on what’s happening here. 6025 minus 4820 equals 𝑐. If we take away 4820 from 6025, we get 𝑐. And that also means that if we add 4820 and 𝑐 together, we would get 6025. And all five of our answer choices are in the form 6025 is equal to something. In our previous stage, we wrote 4820 as 482 tens. And we said that 482 tens was equal to two times 241 times 10. And that means we can say 6025 is equal to two times 241 times 10 plus 𝑐. That looks in some way similar to answer choice a, but it’s missing the times 10. So that’s not an option.

Answer choice b and d are subtracting 𝑐. Those aren’t options. What we need to do is try to regroup this two times 10. We know that two times 10 equals 20. And so, 20 times 241 plus 𝑐 should be equal to 6025. And we can say that 6025 equals 20 times 241 plus 𝑐.

The diagram below shows the second stage of the division algorithm. The number 𝑑 is between one and nine and represents the number of 241s in 1205. What are the values of 𝑑, 𝑒, and 𝑓?

To find 𝑑, we want to know how many times does 241 go into 1205. As we did in stage one, we’re trying to figure out what multiplied by 241 will be as close to 1205 as possible without going over. I know that 200 times five is 1000 and 40 times five equals 200. One times five equals five, which tells me that 241 times five is exactly 1205, which means 𝑑 must be equal to five.

If we plug five in for 𝑑, we then need to multiply five times 241, which we’ve already done over here. That value is substituted in for 𝑒. Five times 241 is 1205, which means 𝑒 equals 1205. And 𝑓 is what we get when we subtract 1205 from 1205. The remainder is zero. 𝑓 is equal to zero. We found that 6025 divided by 241 equals 25.

Our final question was asking the values of 𝑑, 𝑒, and 𝑓, which we found 𝑑 equals five, 𝑒 equals 1205, and 𝑓 equals zero.