Math Module 3 Multi-Digit Multiplication and Division Topic C: Multiplication of up to Four Digits by Single-Digit Numbers Lesson 9: Multiply three- and.

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Math Module 3 Multi-Digit Multiplication and Division Topic C: Multiplication of up to Four Digits by Single-Digit Numbers Lesson 9: Multiply three- and four-digit numbers by one-digit numbers applying the standard algorithm 4.OA.2 4.NBT.5 4.NBT.1 PowerPoint designed by Beth Wagenaar Material on which this PowerPoint is based is the Intellectual Property of Engage NY and can be found free of charge at

Lesson 9 Target You will multiply three- and four-digit numbers by one-digit numbers applying the standard algorithm We are right on target!

Lesson 9 Fluency Expanded Form 343 Say the addition sentence with the answer in standard form = ____

Lesson 9 Fluency Expanded Form 4,679 Say the addition sentence with the answer in standard form. 4, = ____

Lesson 9 Fluency Expanded Form 5,839 Say the addition sentence with the answer in standard form. 5, = ____

Lesson 9 Fluency Expanded Form 528 Say the addition sentence with the answer in standard form = ____

Lesson 9 Fluency Expanded Form 275 Say the number Write the number in expanded form.

Lesson 9 Fluency Expanded Form 4,638 Say the number. 4, Write the number in expanded form.

Lesson 9 Fluency Expanded Form 9,705 Say the number. 9, Write the number in expanded form.

(Write 3 × 2 =.) Say the multiplication sentence in unit form. S: 3 ones × 2 = 6 ones. T: Write the answer in standard form. S: (Write 6.) T: (Write 30 × 2 =.) Say the multiplication sentence in unit form. S: 3 tens × 2 = 6 tens. T: Write the answer in standard form. S: (Write 60.) Lesson 9 Fluency Multiply Mentally Say the multiplication sentence. 432 x 2 = x 2 = ____ 124 x 3 = ____ 312 x 3 = ____ Say the multiplication sentence. 312 x 3 = x 4 = ____ Say the multiplication sentence. 212 x 4 = 848 Say the multiplication sentence. 124 x 3 = x 4 = ____ Say the multiplication sentence. 201 x 4 = x 4 = ____ Say the multiplication sentence. 240 x 4 = 960.

Lesson 9 Fluency Multiply Using Disks On your boards, draw number disks to show this multiplication sentence. 312 x x ____ ones 1 x ____ tens 1 x 312 Fill in the blanks and write the problem vertically. 21 ThousandsHundredsTensOnes  1 x ____ hundreds 3

Lesson 9 Fluency Multiply Using Disks On your boards, draw number disks to show this multiplication sentence. 312 x x ____ ones 2 x ____ tens 2 x 312 Fill in the blanks and write the problem vertically. 21 ThousandsHundredsTensOnes    2 x ____ hundreds 3

Lesson 9 Fluency Multiply Using Disks On your boards, draw number disks to show this multiplication sentence. 312 x x ____ ones 3 x ____ tens 3 x 312 Fill in the blanks and write the problem vertically. 21 ThousandsHundredsTensOnes    3 x ____ hundreds 3

Lesson 9 Fluency Multiply Using Disks On your boards, draw number disks to show this multiplication sentence. 2,154 x ,000 4,308 2 x ____ ones 2 x ____ tens 2 x 2,154 Fill in the blanks and write the problem vertically. 4 5 ThousandsHundredsTensOnes     2 x ____ hundreds 1 2 x ____ thousands 2

Lesson 9 Fluency Multiply Using Disks On your boards, draw number disks to show this multiplication sentence. 212 x x ____ ones 4 x ____ tens 4 x 212 Fill in the blanks and write the problem vertically. 21 ThousandsHundredsTensOnes   4 x ____ hundreds 2

Lesson 9 Fluency Multiply Using Disks On your boards, draw number disks to show this multiplication sentence. 1,504 x ,500 +3,000 4,512 3 x ____ ones 3 x ____ tens 3 x 1,504 Fill in the blanks and write the problem vertically. 4 0 ThousandsHundredsTensOnes    3 x ____ hundreds x ___ thousands

Lesson 9 Application Problem 4 minutes Calculate the total amount of milk in three cartons if each carton contains 236 mL of milk.

Lesson 9 Concept Development Problem 1: Represent and solve 6 x 162 in the place value chart. Relate the process to solving using the standard algorithm. Represent 6 × 162 on your place value chart using the repeated addition way. Work with a partner to solve. Was it necessary to regroup? On my place value chart, I had 6 hundreds, 36 tens, and 12 ones. I regrouped 10 ones for 1 ten and 30 tens for 3 hundreds. My answer is 9 hundreds, 7 tens, and 2 ones.

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 1: Represent and solve 6 x 162 in the place value chart. Relate the process to solving using the standard algorithm. Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help.

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 1: Represent and solve 6 x 162 in the place value chart. Relate the process to solving using the standard algorithm. I multiplied 6 times 2 ones to get 12 ones. We regrouped 10 ones for 1 ten and were left with 2 ones. Tell me what happened in the ones column of your place value chart. Record the number of regrouped tens on the line under the tens column. Record the number of ones in the ones place.

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 1: Represent and solve 6 x 162 in the place value chart. Relate the process to solving using the standard algorithm. I multiplied 6 times 6 tens and got 36 tens. We exchanged 30 tens for 3 hundreds and were left with 6 tens. But we have the 1 ten regrouped from the ones, so 36 tens plus the 1 ten makes 37 tens. So, we have 3 hundreds and 7 tens after we regroup. Tell me what happened in the tens column of your place value chart. Record the number of regrouped hundreds on the line under the hundreds column. Record the number of tens in the tens place. What about the 1 that was written on the line in the tens place, do I need it anymore? We counted it already!

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 1: Represent and solve 6 x 162 in the place value chart. Relate the process to solving using the standard algorithm. We had 6 × 1 hundred = 6 hundreds. 6 hundreds plus the 3 hundreds we regrouped equals 9 hundreds. Now, let’s look at the hundreds. What was the value of the hundreds? Since there’s no need to regroup, write the number of hundreds in the hundreds place. Have we already counted the 3 hundreds we regrouped? What’s the product? Is that the same number we got with the place value chart?

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 2: Solve 5 × 237 using the partial products algorithm. Then solve using the standard algorithm and relate the two methods to each other. Write the expression 5 × 237 vertically on your board. Draw and solve using partial products.

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 2: Solve 5 × 237 using the partial products algorithm. Then solve using the standard algorithm and relate the two methods to each other. Now, let’s solve using the standard algorithm. Starting in the ones column, what do we do? 237 x 5 We multiply 5 times 7 ones and get 35 ones. Tell your partner how you record 35 ones as partial products. 35 ones is 3 tens 5 ones, so we record 3 tens in the tens column and 5 ones in the ones column on the same line. Let’s record 3 tens 5 ones using the standard algorithm. 3 5

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 2: Solve 5 × 237 using the partial products algorithm. Then solve using the standard algorithm and relate the two methods to each other. 237 x Tell your partner what you notice about this recording. The 3 tens is on the line in the tens like in addition and the 5 ones is in the ones place, so it still shows 35 ones. We add partial products together, so the 3 tens on the line means it will get added to the product. Working in the tens column, what do we do next?

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 2: Solve 5 × 237 using the partial products algorithm. Then solve using the standard algorithm and relate the two methods to each other. We multiply 5 times 3 tens and get 15 tens. 237 x tens was recorded on the second line in the partial products method. For the standard algorithm, add 3 tens to 15 tens. 3 tens + 15 tens = 18 tens 8 1 / Say 18 tens as hundreds and tens. 1 hundred 8 tens. Record 1 hundred on the line in the hundreds column and 8 tens in the tens column. Cross out 3 tens because it was added.

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 2: Solve 5 × 237 using the partial products algorithm. Then solve using the standard algorithm and relate the two methods to each other. 237 x hundred + 10 hundreds = 11 hundreds 8 1 / What do we do next? We multiply 5 times 2 hundreds and get 10 hundreds. In the standard algorithm we have to add the 1 hundred that is on the line to make 11 hundreds. Remember to cross of the 1 since we have already included it in our answer. Because there are no more numbers to multiply, we can just record 11 hundreds directly in the product, which is…? / 1,1 1,185

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Lesson 9 Concept Development Problem 2: Solve 5 × 237 using the partial products algorithm. Then solve using the standard algorithm and relate the two methods to each other. 237 x Look back at the work that you did when you solved using partial products. What was the product? It’s the same thing. We came up with the same product even though our methods were different. What are the advantages to the standard algorithm? We record our answer on one line. We are doing all of the calculations in a few steps. 8 1 / / 1,1

Write the expression 6 × 162 again vertically on your board. Let’s find a faster way to express your answer. We will use our place value chart to help. Partial ProductsStandard algorithm Lesson 9 Concept Development Problem 2: Solve 6 x 716 using the partial products algorithm. Then solve using the standard algorithm and relate the two methods to each other. 716 x Let’s try it again! Multiply 6 x 716 using partial products. 9 / 4,2 Now multiply 6 x 716 using the standard algorithm. 716 x ,200 4,296 It’s the same!

Lesson 9 Concept Development Problem 3: Multiply three- and four-digit numbers by one-digit numbers applying the standard algorithm. Shane measured 457 mL of water in a beaker. Olga measured 3 times as much water. How much water did they measure all together? Draw a tape diagram and discuss with a partner how you would solve this problem. We would multiply 4 × 457 to solve this problem. If Olga measured 3 times as much water as Shane, we multiply by four to find the total.

Lesson 9 Concept Development Problem 3: Multiply three- and four-digit numbers by one-digit numbers applying the standard algorithm. Solve using the standard algorithm. What do we multiply first? 4 times 7 ones equals 28 ones. 457 x 4 Show me how to record 28 ones. I write the 2 on the line under the tens place. I write the 8 in the ones. 8 / 2 What do we multiply next? 4 times 5 tens is 20 tens plus 2 tens that were changed from the ones. I have 22 tens. I cross off the 2 because I just included it in the total for the tens. I write 22 tens in my answer. The 2 is written in the hundreds place on the line to show that we regrouped to the hundreds, and the 2 is written in the tens. 2 2

Lesson 9 Concept Development Problem 3: Multiply three- and four-digit numbers by one-digit numbers applying the standard algorithm. What do we multiply next? 4 × 4 hundreds = 16 hundreds. 457 x 4 8 / Then I add 2 hundreds to get 18 hundreds. I cross off the 2 because I included it in the total hundreds. Because there are no more numbers to multiply, I record 18 hundreds directly in my product. / 1,8 They measured 1,828 mL of water all together. The product is 1,828.

Problem Set 10 Minutes

Explain to your partner how you used partial products and the standard algorithm to solve Problems 1(a) and 1(b). Why do both methods work? How are they different?

Look at the questions in Problem 2. Which ones would give you estimates that are very close to the actual product if you rounded the larger number to the hundreds place?

Do you think that you would get a different answer for Problem 4 if the question instead asked you to find 457 times as much as 9? Why or why not?

Explain to your partner how you solved Problem 7. How did you keep track of what each of the numbers meant?

How could you use a tape diagram to represent the work you did on the Application Problem? What significant vocabulary did we use today? Debrief Lesson Objective: Multiply three- and four-digit numbers by one-digit numbers applying the standard algorithm..

Exit Ticket Lesson 1