Lesson 3

Paper-and-Pencil Division

Est. Class Sessions: 2–3

Developing the Lesson

Model Partial Quotients with Area Problems. In this lesson, students review a paper-and-pencil method, the partial quotients method, for solving division problems. We teach this method because many students find it easier to learn than the traditional method taught in the U.S. Both methods involve making estimates about the quotient, but the partial quotient method is more flexible. It allows students to underestimate and to work with easier numbers.

For a more detailed discussion of the partial quotients method, see the Mathematics in this Unit and the Letter Home.

  • Professor Peabody wants to paint the floor of a hall that is 8 feet wide. The label on his paint can says he has enough to cover 280 square feet. How far down the hallway can Professor Peabody paint with his can of paint?

Draw a picture of the hallway as in Figure 1 and encourage them to solve the problem by making and refining estimates as they did before. They can record their calculations any way they want. Explain that after they solve the problem, you will talk about an organized way to record their calculations.

After students have solved the problem, ask a volunteer to share his or her solution. As the student describes the estimates in his or her solution, show how to record them in the partial quotients method. Figures 2–4 show each step as a student might write it using the rectangle model and the comparable step as written in the format of the partial quotients method.

In Figure 2, the first estimate is 10 feet. Multiply 8 × 10 = 80 shows that after painting 10 feet down the hall, 80 square feet of floor have been painted. Write 80 under 280 and subtract. There is enough paint remaining to cover another 200 square feet.

Figure 3 shows a second estimate of 20 feet. Multiply 8 × 20 = 160, write the product under the 200, and subtract. There is enough to cover 40 more square feet.

Figure 4 shows a third estimate of 5 feet. Multiply 8 × 5 = 40, write the product under 40, and subtract to get zero. Stop here. All the paint has been used. Adding the numbers on the right, the partial quotients, you see that 10 + 20 + 5 = 35 feet of hall were painted. We write the quotient 35 on top of the original problem so it can be easily seen.

280 ÷ 8 = 35

Compare Methods.

  • What do you notice that is the same about the two methods? (In both methods, estimates are made and then added to find the answer. After each estimate, the estimate is multiplied by the divisor and then the product is subtracted from the dividend to determine how much is left to be divided; then the steps are repeated with what is left.)
  • What do you notice that is different? (The way it is written looks different; the rectangle model shows the multiplication step and the partial quotient method shows the subtracting step.)
  • Can the same estimates be used with both methods? (yes)
  • Do the two methods reach the same answer? (yes)

Model Partial Quotient with Sharing Problems. Another way to think about division is with a sharing problem.

  • A box contains 644 beads. If the beads are to be shared equally among 7 children, how many beads will each child get?

Ask students if they recall a similar problem from Lesson 2. They should recall that in the opening problem, a collection of rocks was shared among 5 students.

  • What method did we use to solve that division problem? (We drew columns to keep track of the rocks as we were dividing them up.)
  • Would that same method work to solve this problem? Why do you say so? (Yes, because in both problems we are taking a group of things and sharing them equally among a number of people.)
  • How many columns should I draw for this problem? Why? (7 columns because the beads are divided among 7 children.)
  • Will each child get more than 100 beads? Tell me your thinking. (No, 100 is too high since 7 × 100 = 700.)
  • Estimate. What is a reasonable guess for the number of beads each child will get?

Write student estimates on the board or a display to be checked after the problem has been solved completely.

Solve a Problem Using the Column and the Partial Quotients Methods. Ask students what amount they would choose for an initial distribution of the beads. Suggest using small, convenient numbers (numbers ending in zeros are good) at the beginning. Figures 5 and 6 show 20 as a first estimate, then 50, then 20 again, and finally 2.

As you work through the problem with the class, have a student write the estimates and calculations in the columns as you write them in the partial quotients format. Ask students to suggest possible estimates and ask them to explain what each number in the problem represents.

  • [Student name] has suggested 20 for the first estimate. What does that 20 mean? (The 20 means that we give each child 20 beads.)
  • Where did we write the 20 in the column method? (We write a 20 in each of the columns to show that each child got 20 beads.)
  • Where did we write the 20 in the partial quotients method? (to the right of the problem)
  • How many beads total have we shared so far? How do you know? (140 beads; We gave 20 beads each to 7 children. 7 × 20 = 140.)
  • Where did we write the 140 beads in the columns method? (under the words “Into the Columns”)
  • Where did we write it using the partial quotients method? (under the dividend, 644)
  • How many beads do we have left that still have to be shared? How do you know? (We have shared 140 so we subtract: 644 beads to start minus 140 beads = 504 beads left to share.)
  • Where do we write that in our two methods? (In the column method, it is written under “Left to Divide.” In the partial quotients method, the subtraction is shown under the dividend.)
  • We have 504 beads left to be shared. How many should we distribute in the next step? Who can suggest another estimate?
  • Estimate the number of times 7 goes into 504.

In this example we estimate that 7 goes into 504 fifty times. Then multiply 50 times 7 to get 350 and subtract. There are now 154 beads left to be distributed. We continue in this manner until no more beads can be distributed.

  • What does the last number sentence, [14 − 14] = 0, tell us? (It tells us that we have no more beads. We have shared them all.)
  • How many shares or equal groups did we make? How many children received beads? (7 children received beads.)
  • How many beads did each child get? How do you know? (Each child got 92 beads; I know because I added up how many we gave in each step. 20 + 50 + 20 + 2 = 92.)
  • Were there any beads left over? How do you know? (There were no beads left over because at the end we had 14 beads and gave 2 to each child and that took care of all the beads. 7 × 2 = 14; 14 − 14 = 0 beads.)

Compare Methods. After finishing the problem, write the quotient (92) above the dividend.

  • How are the two methods alike? (They both have the same steps: making an estimate, subtracting the product of the estimate and the divisor from the dividend, making a new estimate, etc., until there is nothing left of the dividend to be divided; then all the estimates are added to find the answer.)
  • How are the two methods different? (Using the columns, we write the estimates from the bottom of the column and move up; with the partial quotients, the work is from top to bottom. The column method has a lot written, but the partial quotients method has only the partial estimates written on the side.)
  • How can you check your answer to see if it is correct? (Use multiplication.)

Students can check the answer by multiplying to see whether the quotient times the divisor gives the dividend. That is, to check that 644 ÷ 7 = 92, check that 92 × 7 = 644.

Use Partial Quotients. Use the partial quotients division method together with several more problems:

448 ÷ 8 (Answer = 56)
985 ÷ 5 (Answer = 197)
3456 ÷ 9 (Answer = 384)

For each problem, ask students to estimate their answers first. For example, 448 ÷ 8 is less than 60 since 8 × 60 = 480. It is more than 50 since 8 × 50 = 400. For the second problem, they can easily see the answer is more than 100 since 5 × 100 = 500 and it is less than 200, since 5 × 200 = 1000.

While the class is solving the problems, look for students who solve a problem using different estimates and therefore a different set of steps to complete the problem. Ask the selected students to record their solutions on the board side by side. See Figure 7.

  • What do you notice about the steps in the solutions? (In the examples shown, one solution has 4 partial quotients, one has 3, and one has 2.)
  • Why was one student able to do the problem in fewer steps? (He or she selected larger estimates.)
  • Which strategy seems most efficient? (Students may select the strategy with the least steps. Some students will choose the strategy where they can do the multiplication easily in their heads. Either response is acceptable.)

Solve Problems with Remainders.

  • A large grocery store baked 2349 cookies. They sold the cookies in packages of 8. How many packages of cookies did they make?
  • Will the number of packages be over 100? Over 500? (There will be more than 100 packages because 8 × 100 = 800. There will be less than 500 packages because 8 × 500 = 4000.)

Ask students to solve the problems and share one or two solutions with the class. A sample solution is shown in Figure 8.

The discussion prompts that follow are adapted from a classroom discussion in which the teacher checked for students' understanding of the steps in the partial quotients method. These questions refer to the solution in Figure 8.

  • What does the 400 mean in this problem? What does it mean to you? (It means that 50 packages of 8 cookies each have been made, so 400 cookies have been put in packages.)
  • Where did the 1949 come from? (from subtracting the 400 cookies from the 2349 cookies we started with)
  • What does that number mean in this problem? (There are 1949 cookies left to package.)
  • What was the next partial quotient? (100)
  • What does that mean in this problem about packaging cookies? (It means that we have made another 100 packages)
  • What does the 800 mean? (that 800 more cookies have been put in 100 packages)

Note that the problem in Figure 8 has a remainder.

  • How many total packages did they make? (293 packages)
  • What does the 5 tell you? (There are 5 cookies remaining that could not be packaged.)
  • How can you check the answer with multiplication? (Since 2349 ÷ 8 = 293 with remainder 5, then 293 × 8 + 5 = 2349.)

Students can quickly check this calculation on a calculator, or they can check with paper-and-pencil multiplication.

Practice the Partial Quotients Method. The Paper-and-Pencil Division pages in the Student Guide provide additional explanation and practice. Have students read and discuss the two examples in pairs. Each partner can explain one example to the other. The first example has a whole number quotient. The second has a remainder. Ask questions to check that students understand what the numbers mean in each step.

Have students complete Question 1 in groups. Ask them to explain the remainders. In Question 2, students check their answers using multiplication.

Question 3 presents four solutions to the same division problem. By now, students will have noticed that for each problem, there are several ways to choose partial quotients but they all result in the same answer. This makes the method very flexible and allows students to choose numbers they are comfortable working with. However, some choices are more efficient than others. When students are ready, they can start to look for choices that solve the problem in fewer steps.

  • What is the first estimate in each solution in Question 3? (20, 100, 125, 140)
  • Which one would you have tried first? Why? (Possible responses include: I would pick 100 because 5 × 100 is easy and a big number that gets closer to 700. I picked 125 × 5 because it is easy to multiply when I think about money. I know 5 quarters is 125 added to 500 is 625.)
  • How do the estimates affect the next steps in the solutions? (If you can think of larger estimates, you can do the division in fewer steps.)

In general, when students choose partial quotients as large as possible (without going over), they will solve the problem in the fewest number of steps. In this case, the fourth solution has the same number of steps as the traditional long division method.

Have students complete Questions 4–5 in small groups or pairs. Have them compare their solutions and note the number of steps they used.

When students have completed Questions 1–5 in groups or pairs, have them work independently to answer Check-In: Questions 6–10. As they work, circulate and observe which students are ready to look for larger first estimates to solve the problem in fewer steps and which students are not. Encourage, but do not require, students to solve the problem using the fewest steps. Encourage students to choose numbers that are comfortable for them, even if it takes more steps.

Use students' responses to Check-In: Questions 6–10 in the Student Guide to assess students' abilities to show connections between models and strategies for multidigit division [E2]; interpret remainders from division of multidigit numbers [E3]; divide numbers that are multiples of ten [E6]; and divide multidigit numbers by one- and two-digit divisors using paper and pencil [E7].

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Setting up the problem
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First estimate: 10 ft. That leaves 200 sq. ft. to paint.
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Second estimate: 20 ft. That leaves 40 sq. ft. to paint.
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Third estimate: 5 ft. No paint remains.
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Partial quotients method for 644 ÷ 7
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Column method for 644 ÷ 7
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Three ways to use partial quotients to solve 985 ÷ 5
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Computing 2349 ÷ 8 using the partial quotients method
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