Warm-up: Number Talk: Three Factors (10 minutes)
The purpose of this Number Talk is to elicit strategies and understandings students have for multiplying three factors, one of which is ten. These understandings help students develop fluency and will be helpful when students apply the standard algorithm to find the product of a three-digit and a two-digit number.
Students have an opportunity to look for and make use of structure (MP7) because they can use the distributive property to find a product using previous calculations.
- Display one expression.
- “Give me a signal when you have an answer and can explain how you got it.”
- 1 minute: quiet think time
- Record answers and strategy.
- Keep expressions and work displayed.
- Repeat with each expression.
Find the value of each product mentally.
- \((2 \times 3) \times 10\)
- \((2 \times 40) \times 10\)
- \((2 \times 200) \times 10\)
- \((2 \times 243) \times 10\)
- “How did multiplying all the products by 10 influence the result?” (It made the result ten times as big, so the digits all shift one place to the left and it has a zero at the end.)
- “How are the products \(2 \times 243\) and \(20 \times 243\) related?” (The second one is ten times as big, so the digits shift one place to the left and it has a 0 at the end.)
- “You can use this idea today when we apply the standard algorithm to find products of a 3-digit number and a 2-digit number.”
Activity 1: Compose a New Unit (25 minutes)
The goal of this activity is to use the standard algorithm to find products in which composition of a new unit happens once. Students first calculate a 3-digit and 2-digit example using a strategy of their choice and then analyze the same example done with composition recorded above the product. Students may use different strategies when they try on their own including
- partial products
- mentally accounting for the hundred that is composed when finding the product \(3 \times 40\)
After students discuss how composing new units is recorded in the algorithm, they find the value of two multiplication expressions using the standard algorithm.
When students interpret a new way of multiplying a 3-digit and 2-digit number, they use their understanding of place value to make sense of the method (MP7).
Advances: Speaking, Conversing, Representing
- Groups of 2
- “You are now going to learn how to compose and record new units for a three-digit and two-digit product.”
- “Work with your partner on the first 2 problems.”
- 2-3 minutes: independent work time
- 5-7 minutes: partner work time
- Find the value of \(241 \times 23\) .
- Lin used the standard algorithm to find the value of \(241 \times 23\). Here is her work:
- Where do you see \(241 \times 3\) in Lin’s work?
- Where do you see \(241 \times 20\) in Lin’s work?
- What does the 1 above 241 represent in Lin’s calculation?
- Use the standard algorithm to find the value of \(182 \times 41\).
- Use the standard algorithm to find the value of \(304 \times 23\).
- Invite students to share how they interpret Lin’s work finding \(241 \times 23\).
- Display the image of Lin’s calculation.
- Circle the 2 in the number 723 in Lin’s calculation.
- “What does the 2 in the tens place represent?” (It’s 2 of the tens from \(3 \times 40\).)
- “What does Lin do with the other 10 tens?” (She makes a hundred out of them and puts them together with the other hundreds when she multiplies 200 by 3.)
- Circle the 1 above 241 in Lin's work.
- "What does this 1 represent?" (It’s the hundred from \(3 \times 40\).)
- Circle the partial product 4,820.
- “What does 4,280 represent in the calculation?” (\(20 \times 241\). The 2 from the factor 23 is in the tens place and so it represents 20.)
- “Now take a few minutes to solve the last two problems.”
- 4-5 minutes independent work time
- Invite students to share the products and ask students what questions they have about the standard algorithm with composition.
Activity 2: All the Products (10 minutes)
The goal of this activity is to multiply numbers with no restrictions on the number of new units composed. Students first multiply a 3-digit number by a 1-digit number and a 3-digit number by a 2-digit number with no ones. They can then put these two results together to find the product of a 3-digit and 2-digit number with many carries. They then solve one more 3-digit and 2-digit example with no scaffold. Because these calculations have new units composed in almost every place value, students will need to locate and use the composed units carefully. It gives students a reason to attend to the features of their calculation and to use language precisely (MP6).
Supports accessibility for: Organization, Conceptual Processing
- “You are going to find products with many new composed units. As you work, think carefully about where you place these values.”
- 8–10 minutes: independent work time
- 3–5 minutes: partner discussion
- Monitor for students who use the results of the first two calculations to find the third, and for students who correctly compose all the new place values.
Find the value of each product using the standard algorithm.
- \(647 \times 9\)
- \(647 \times 50\)
- \(647 \times 59\)
- \(264 \times 38\)
- Display the expression: \(647 \times 59\).
- “How did you use the first two calculations to help with the third problem?” (They gave me the two partial products for the product \(647 \times 59\), so I just had to add them up.)
- Invite students to share their responses for the last product, focusing on the newly composed units.
“Today, we practiced using the standard algorithm to multiply multi-digit numbers with new units composed.”
“What do you have to think about when you are multiplying and a lot of new units are composed?” (You have to keep track of how you record the units. You can make an estimate to see if your answer is reasonable.)
Display student work for \(264 \times 38\) from activity 2 or use the example from the student responses.
"Where did we compose new units when we solved this problem?" (When we multiplied to find the two partial products, we had to compose new units above the 2 and 6 in 264. When we added the partial products, we composed a new one thousand above the 2.)
"How is composing new units when we multiply the same as composing new units when we add?" (When I am multiplying or adding numbers sometimes I get a value that's too much for the place I'm in. The composed units are recorded separately and then I add them.)
"How is composing new units when we multiply different from composing new units when we add?" (When we multiply, we multiply and then add the new units. When we add, we are adding the whole time, there is no multiplication.)