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

Math Module 3 Multi-Digit Multiplication and Division

Topic C: Multiplication of up to Four Digits by Single-Digit NumbersLesson 9: Multiply three- and four-digit numbers by one-digit numbers applying the standard

algorithm

4.OA.2 4.NBT.5 4.NBT.1PowerPoint designed by Beth Wagenaar

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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

FluencyExpanded Form

343

• Say the addition sentence with the answer in standard form.

300 + 40 + 3 = ____

Lesson 9


4,679


4,000 + 600 + 70 + 9 = ____

Lesson 9


5,839


5,000 + 800 + 30 + 9 = ____

Lesson 9


528


500 + 8 + 20 = ____

Lesson 9


275• Say the number.

200 + 70 + 5• Write the number in expanded form.

Lesson 9


4,638• Say the number.

4,000 +600 + 30 + 8

• Write the number in expanded form.

Lesson 9


9,705• Say the number.

9,000 +700 + 5

• 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 9FluencyMultiply Mentally

• Say the multiplication sentence.

• 432 x 2 = 864.

432 x 2 = ____ 124 x 3 = ____

312 x 3 = ____• Say the multiplication

sentence. • 312 x 3 = 936.


sentence. • 212 x 4 = 848

• Say the multiplication sentence.

• 124 x 3 = 372.


sentence. • 201 x 4 = 804.


sentence. • 240 x 4 = 960 .

Lesson 9FluencyMultiply Using Disks

On your boards, draw number disks to show this multiplication sentence.

312 x1 2

10+300 312

1 x ____ ones

1 x ____ tens

1 x 312Fill in the blanks and write the problem vertically.

21

Thousands Hundreds Tens Ones

=== = ==

1 x ____ hundreds

3



312 x2 4

20+600 624

2 x ____ ones

2 x ____ tens


21


===

===

=

=

==

==

2 x ____ hundreds

3



312 x3 6

30+900 936

3 x ____ ones

3 x ____ tens


21


=========

===

======

3 x ____ hundreds

3



2,154x 2 8100200

+4,000 4,308

2 x ____ ones

2 x ____ tens

2 x 2,154Fill in the blanks and write the problem vertically.

45


==

==

=

=

=====

=====

====

====

2 x ____ hundreds

12 x ____ thousands

2



212 x4 8

40+800 848

4 x ____ ones

4 x ____ tens


21


========

====

========

4 x ____ hundreds

2



1,504 x3

120

1,500+3,000 4,512

3 x ____ ones

3 x ____ tens

3 x 1,504Fill in the blanks and write the problem vertically.

40


= = =

===============

============

3 x ____ hundreds

513 x ___ thousands

Lesson 9Application Problem4 minutes

Calculate the total amount of milk in three cartons if each carton contains 236 mL of milk.

Lesson 9Concept DevelopmentProblem 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.


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.




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.




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!




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?


Lesson 9Concept DevelopmentProblem 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.




• Now, let’s solve using the standard algorithm. Starting in the ones column, what do we do?

237x 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.

35




237x 535

• 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?




• We multiply 5 times 3 tens and get 15 tens. 237x 53

5• 15 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 tens8

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.




237x 53

51 hundred + 10 hundreds = 11 hundreds

81 /

• 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




237x 53

5

• 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.

81 //

1,1


Partial Products Standard algorithm

Lesson 9Concept DevelopmentProblem 2: Solve 6 x 716 using the partial products algorithm. Then solve using the standard


716x 63

6

• Let’s try it again! Multiply 6 x 716 using partial products.

9/

4,2

• Now multiply 6 x 716 using the standard algorithm.

716x 6

3660

+4,2004,296 It’s the same!

Lesson 9Concept DevelopmentProblem 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.


the standard algorithm.• Solve using the standard algorithm. What do we multiply first?• 4 times 7 ones equals 28 ones.

457x 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


the standard algorithm.• What do we multiply next? • 4 × 4 hundreds = 16 hundreds.

457x 4

8/22

2

• 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 Set10 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?

DebriefLesson Objective:Multiply three- and

four-digit numbers by one-digit numbers

applying the standard algorithm. .

Exit Ticket Lesson 1

Documents

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