Lesson 19
Division With and Without Remainders
Warm-up: Notice and Wonder: Equations with Hundreds (10 minutes)
Narrative
This warm-up prompts students to analyze patterns and look for structure in division equations (MP7), and to reinforce their understanding of factors and multiples.
Launch
- Groups of 2
- Display the equations.
- “What do you notice? What do you wonder?”
- 1 minute: quiet think time
Activity
- “Discuss your thinking with your partner.”
- 1 minute: partner discussion
- Share and record responses.
Student Facing
What do you notice? What do you wonder?
\(\begin{align}100 &= 33 \times 3 + 1\\ \\ 200 &= 66 \times 3 + 2\\ \\ 300 &= 100 \times 3\\ \\ 400 &=133 \times 3 +1\\ \\ 500 &=166 \times 3 +2\\ \\ 600 &= 200 \times 3 \end{align} \)
Student Response
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Activity Synthesis
- “If we continue the pattern through 1,000, what would the equations look like?”
- “In all the equations, there is multiplication by 3, which suggests we could think of them in terms of division by 3.”
- “How could we interpret \(300 = 100 \times 3\) in terms of division by 3?” (Dividing 300 by 3 gives 100.)
- “How about \(600 = 200 \times 3\)?” (Dividing 600 by 3 gives 200.)
- “Can we interpret the equation \(100 = 33 \times 3 + 1\) in terms of division by 3? What does it tell us?” (Yes. Dividing 100 by 3 gives 33 and a remainder of 1.)
- “What about \(500 = 166 \times 3 + 2\)?” (Dividing 500 by 3 gives 166 and a remainder of 2.)
Activity 1: A Stack of Partial Quotients (15 minutes)
Narrative
This activity develops students’ understanding of the vertical method of recording partial quotients and their ability to use it to perform division. One of the quotients here involves a remainder, prompting students to interpret it. Students are reminded that the term “remainder” is used to describe “leftovers” when dividing.
Advances: Conversing, Reading
Launch
- Groups of 2
Activity
- 3 minutes: independent work time on the first 2 problems.
- Pause after problem 2 to discuss students’ responses.
- Display the different ways that students decompose 389 to divide it by 7.
- “Most other calculations we’ve seen so far end with a 0, but this one ends with a 4. What does the 4 tell us?” (We cannot make a group of 7 with 4 leftover. 389 is not a multiple of 7, and there are leftovers.)
- “When we divide and end up with leftovers we call them remainders, because they represent what is remaining after we divide into equal groups.”
- Display: \(389 = 7 \times 55 + 4\)
- “How does this equation show that \(389 \div 7\) has a remainder?” (It shows that 389 is not a multiple of 7. It also shows that 7 and 55 make a factor pair for 385, and 389 is 4 more than that.)
- 3 minutes: independent work time on the last 2 problems.
- As students work on the last two problems monitor for students who:
- start with the largest multiple of 3 and 10 within 702 that they can think of to decompose the dividend (690, 600).
- use the fewest steps to find the quotient.
Student Facing
Jada used partial quotients to find out how many groups of 7 are in 389.
Analyze Jada’s steps in the algorithm.
-
- Look at the three numbers above 389. What do they represent?
- Look at the three subtractions below 389. What do they represent?
- What is another way you can decompose 389 to divide by 7?
- Is 389 a multiple of 7? Explain your reasoning.
- Use an algorithm that uses partial quotients to find out how many groups of 3 are in 702.
- Is 702 a multiple of 3? Explain your reasoning.
Student Response
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Activity Synthesis
- Display responses that show variation in how students decomposed 702 in problem 3.
- Highlight that there are countless ways of using partial quotients to divide a number, but it may be more efficient to divide larger groups than smaller groups (for example, removing 350 once, as opposed to removing 70 five times. Removing smaller groups is just as valid, however.)
- To help students see the relationship between corresponding partial products and partial quotients, consider illustrating them visually. Some examples:
Activity 2: Andre and Elena’s Work (10 minutes)
Narrative
In this activity, students apply their understanding of partial quotients and the vertical recording method to divide four-digit numbers. They also identify some errors that are common when finding quotients this way. When students determine where the errors are and correct them, they critique the reasoning of others and construct viable arguments (MP3).
Supports accessibility for: Visual-Spatial Processing, Organization
Launch
- Groups of 2
Activity
- 1 minute: quiet think time for the first question
- Briefly discuss student responses. Highlight that multiples of 5 end with 0 and 5, so 2,316 is not a multiple of 5 and the division will result in a remainder.
- 3 minutes: independent work time for the second question.
- 1–2 minutes: partner discussion
Student Facing
Andre and Elena are dividing 2,316 by 5. Before they begin, Andre says, “I can already tell that there will be a remainder.”
- Without doing any calculations, decide if you agree with Andre. Explain your reasoning.
-
Here is Andre and Elena’s work. Each student made one or more errors. Identify the errors each student made. Then, show a correct computation.
Andre's Work
Elena's Work
Student Response
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Activity Synthesis
- Select students to share their responses to the last question.
- “What are some ways to check our answers and avoid the mistakes Andre and Elena made? For example, how can we tell if dividing 2,316 by 5 gives a result in the 100s or in the 400s?” (We can estimate by multiplying the result by 5: \(5 \times 103\) is a little over 500, and \(5 \times 400\) is 2,000.)
- To reinforce the importance of keeping track of the partial quotients during division, consider annotating a corrected version of Elena's algorithm. Record the product that corresponds to each number being subtracted from the dividend (the 1,500, 500, 300, and 15).
Activity 3: Incomplete Calculations (10 minutes)
Narrative
Launch
- Groups of 2–4
- “Choose at least two calculations to finish. Make sure each calculation is completed by someone in your group.”
Activity
- 3–4 minutes: independent work time
- 2 minutes: group discussion
Student Facing
Here are four calculations to find the value of \(3,\!294 \div 3\), but each one is unfinished.
Complete at least two of the unfinished calculations. Be prepared to explain why you chose them.
A
B
C
\(\begin{align} 600\div 3&= \phantom{000000}\\ 600\div 3&=\\ 600\div 3&= \\600\div 3&= \\600\div 3&= \\270\div 3&= \end{align}\)
D
\(\begin{align} 3,\!300\div 3&= 1,\!100\\-\hspace{10mm}6\div 3&= \phantom{0000\!}2\\ \overline {\hspace{5mm}\phantom{3,\!294 \div 3}} &\overline{\phantom{0000000000}}\end{align}\)
Student Response
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Activity Synthesis
- “How are the four strategies alike? How are they different?” (The first three are similar in that they involve partial quotients. The last one involves estimation.)
- “Which strategy or strategies do you find easy to follow? Hard to follow?”
- “Which strategy seems the most efficient? The least efficient?” (Calculations B and C seem lengthy and could be shortened by using larger multiples of 3. The last seems most efficient.)
Lesson Synthesis
Lesson Synthesis
“Today we looked at different ways to divide multi-digit numbers by one-digit divisors. Some divisions result in a number with a remainder and others result in no remainders.”
“Can we always tell if there will be a remainder?” (No, not always, but sometimes we can.)
“How can we sometimes tell that there will be a remainder?” (We can use what we know about the multiples of a number. For example, all multiples of 2, 4, 6, and 8 have an even number for the last digit. All multiples of 5 end in 5 or in 0.)
“Some ways to divide are pretty lengthy. What are some ways to divide efficiently?” (Dividing larger portions of the dividend, or taking larger multiples of the divisor.)
“In the last activity, we saw estimating as a rather efficient way to find a quotient. How might we use estimation to find \(5,\!970 \div 3\) or \(6,\!986 \div 7\)?” (Notice that:
- \(5,\!970\) is 30 less than \(6,\!000\), which is \(3 \times 2,\!000\). Thirty is \(3 \times 10\), so \(5,\!970\) is \(3 \times 1,\!990\).
- \(6,\!986\) is close to \(7,\!000\), which is \(7 \times 1,\!000\), and \(6,\!986\) is 14 or \(7 \times 2\) less than \(7,\!000\). So \(6,\!986\) is \(7 \times 998\).)
“How can we check the result of our division to make sure it’s not off?” (We can multiply the result by the divisor, adding the remainder if there is one, and see if it gives the dividend.)
Cool-down: Find a Quotient (5 minutes)
Cool-Down
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