Mission 3 Summary

Learning Objectives

Students will use area and perimeter formulas for rectangles to solve word problems involving comparisons. They'll show their understanding by solving multi-step real-world problems.
Students will explore patterns in multiplying by 10, 100, and 1,000 using arrays and numbers. They will use the area model to multiply larger multiples of 10 and show their understanding with visual and numerical methods.
Students will use place value disks to model two-digit by one-digit multiplication, then extend to larger numbers. They will connect the area model and partial products to the standard algorithm for a deeper understanding of multiplication.
Students will solve two-step word problems involving multiplication and division comparisons. They will explain their reasoning and check their answers by justifying their problem-solving process.
Students will solve division word problems with and without remainders using arrays, area models, and place value disks. They'll break down two-digit dividends, decompose remainders, and explain their solutions using place value and models.
Students will identify and list factor pairs for numbers up to 100. They will use division and the associative property to verify these pairs and recognize patterns in factorization.
Students will divide multiples of 10, 100, and 1,000 by single digits, solve four-digit division problems using place value disks, and handle zeros in dividends or quotients. They will interpret division word problems, find quotients and remainders, and connect the area model to long division.
Students will multiply two-digit multiples of 10 using a place value chart and the area model, then move to multiplying two-digit numbers with partial products. They will transition from this method to the standard algorithm for two-digit multiplication.

Area and Perimeter:

Students explore the formulas for area and perimeter, solving multiplicative comparison problems that involve the language of "times as much," with a focus on problems using area and perimeter as a context.

Formulae:

Area of rectangle = l x b

Perimeter of rectangle = 2(l + b)

For example:

Area = l X b

= 12 X 8 ft

= 96 square ft.

Perimeter = 2 X (l + b)

= 2 X (12 + 8)

= 2 X ( 20 )

= 40 ft.

Units of Area and Perimeter:

Understanding Place Value in Multiplication:

Students use place value disks to multiply single-digit numbers by multiples of 10, 100, and 1,000, as well as two-digit multiples of 10 by two-digit multiples of 10, reasoning between arrays and written numerical work to understand the role of place value units in multiplication,

Example 1: 40 x 3

Example 2: 400 x 3

Example 3: 4000 x 3

Mastering Multiplication Through Decomposition and the Distributive Property

Students decompose numbers into base ten units to find products of single-digit by multi-digit numbers, using the distributive property and place value disks to model and solve multiplication problems.

They practice recording partial products and bridge these to the standard algorithm, while also comparing and connecting the partial products method, the standard algorithm, and the area model through the distributive property.

Example: 1423 x 3

Using Place Value Disc

Steps

1. Write down the multiplication problem you want to solve.

2. Set Up the Place Value Discs for 1423:

a. Thousand (1): Place 1 disc in the thousands column.

b. Hundreds (4) : Place 4 discs in the hundreds column.

c. Tens (2) : Place 2 discs in the tens column.

d. Ones (3) : Place 3 discs in the ones column.

3. Multiply Each Place by 3:

a. Thousands: 1 × 3 = 3 discs in the thousands place.

b. Hundreds: 4 × 3 = 12 discs in the hundreds place.

c. Tens: 2 × 3 = 6 discs in the tens place.

d. Ones: 3 × 3 =9 discs in the ones place.

4. Regroup If Necessary

In the hundreds place, you have 12 discs. Since 10 hundreds equal 1 thousand, you regroup:

a. Move 10 discs from the hundreds place to the thousands place, which adds 1 disc to the thousands place.

b. You are left with 2 discs in the hundreds place.

Now, the thousands place will have 3 original discs plus 1 regrouped disc, making a total of 4 discs.

5. Write Down the Final Answer:

a. Thousands Place: 4 discs.

b. Hundreds Place: 2 discs (after regrouping).

c. Tens Place: 6 discs.

d. Ones Place: 9 discs.

So, 1423×3=4269

Area Model

Steps

1. Break Down the Numbers

1423 can be broken down into its place values:

1000
400
20
3

2. Set Up the Area Model

Draw a large rectangle and divide it into four smaller sections, one for each part of the number 1423. Label each section with the corresponding place value of 1423.

3. Multiply Each Part

Now, multiply each part of the number 1423 by 3 to get partial products.

1000 × 3 = 3000
400 × 3 = 1200
20 × 3 = 60
3 × 3 = 9

Update your area model with these products:

4. Add the Products

Finally, add up all the partial products to get the final result.

3000 + 1200 + 60 + 9 = 4269

So, 1423 × 3 = 4269 using the Area Model.

Word Problems:

Q: Ron is helping his mother plant flowers in the garden. On the first day, they planted 75 flowers. On the second day, they planted 3 times as many flowers as on the first day. However, due to heavy rain, 25 flowers were damaged on the second day.

1. How many flowers did they plant on the second day?

2. How many flowers did they plant altogether on both days?

3. After the damage, how many healthy flowers were left?

Solution:

Steps:

1. Multiply:

On the second day, they planted 3 times as many flowers as on the first day:

75×3=225 flowers.

2. Add:

Total flowers planted on both days:

75 + 225 = 300 flowers.

3. Subtract:

After the damage, subtract the damaged flowers from the total planted on the second day:

225 − 25 = 200 healthy flowers on the second day.

Total healthy flowers after two days:

75 + 200 = 275 flowers.

This problem gives students the chance to write equations from the word problem and solve it step-by-step using multiplication, addition, and subtraction.

Exploring Factor Pairs and Patterns in Numbers

Find all the factor pairs of 72.

1. List all the factor pairs of 72.

2. Verify the factor pairs using division.

3. Can you identify any patterns in the factor pairs?

Steps:

1. List the factor pairs of 36:

Start by finding numbers that divide 36 without leaving a remainder.
Factor pairs:

1 × 36

2 × 18

3 × 12

4 × 9

6 × 6

2. Verify using division:

36 ÷ 1 = 36,

36 ÷ 2 = 18

36 ÷ 3 = 12

36 ÷ 4 = 9

36 ÷ 6 = 6

3. Identify patterns:

As the factors increase on one side, the corresponding factor decreases on the other.
Each factor pair multiplies to give the original number (36).

This exercise helps students identify factors and recognize patterns while reinforcing the concept of division and factorization.

Applying the Distributive Property to Multiply Larger Numbers

The distributive property allows you to break down a multiplication problem into smaller, more manageable parts.

Steps:

1. Break the number 84 into two parts:

You can split 84 into 80 and 4, because

84 = 80 + 4

2. Apply the distributive property:

Instead of multiplying 84 directly by 3, you can multiply each part separately:

84 × 3 = (80+4) × 3

Using the distributive property, this becomes:

(80 × 3) + (4 × 3)

3. Multiply each part:

80 × 3 = 240

4 × 3 = 12

4. Add the results:

240 + 12 = 252

So, 84 × 3 = 252

The distributive property lets you split a number into parts, multiply those parts separately, and then add them together to get the same result. It's especially helpful for doing mental math with larger numbers.

Understanding Divisibility: When is a Number Divisible?

A number is said to be divisible by another number when it can be divided evenly, with no remainder.

For example:

72 ÷ 3 = 24 (no remainder) → 72 is divisible by 3.

When a number is not divisible, it leaves a remainder.

For example:

25 ÷ 3 = 8 with a remainder of 1 → 25 is not divisible by 3.

Divisibility means the division results in a whole number, while non-divisibility leaves a remainder.

Exploring Unknown Group Sizes and Place Value Connections

Students apply their new multiplication skills to solve multi-step word problems and comparison problems by writing equations and using addition, subtraction, and multiplication to find the answers.

They build on their Grade 3 knowledge of division (where the group size or number of groups is unknown) and deepen their understanding of place value.

Example: 240 ÷ 3

Steps:

Step 1: Set Up the Place Value Chart

Draw a place value chart with two columns: one for Tens and one for Ones.

Step 2: Represent the Number 24

Place the number 24 into the chart according to its place values:

2 in the Tens column (representing 20)
4 in the Ones column

Step 3: Divide Each Place Value by 3

Now, divide each digit by 3.

The 2 (which is 20) cannot be divided completely by 3, so it needs to be regrouped into the Ones column.
Ones: Add the 2 tens (or 20) to the 4 ones, making it 24 ones.
- 24 ÷ 3 = 8

Step 4: Place the Result in the Chart

Since there’s no complete group of tens (because 2 divided by 3 gives a fraction), the Tens column will have 0.

The Ones column will have 8, since 24 divided by 3 equals 8.

Question: Solve 96 ÷ 4

Students start by representing division with single-digit divisors using arrays and the area model, then practice with place value disks.
They learn the standard division algorithm by breaking down units step by step. Finally, they use the area model to solve division problems, both with and without remainders.

Example: 96 ÷ 4:

Steps:

1. Set Up the Division:

Write 96 inside the division bracket (the dividend) and 4 outside the bracket (the divisor).

2. Divide the First Digit:

Look at the first digit of 96, which is 9. Determine how many times 4 can go into 9 without exceeding it.

4 × 2 = 8 , so 4 goes into 9 two times.

Write 2 above the division bracket, aligned with the 9.
Multiply 2 by 4 and write the result (8) below 9.

3. Subtract and Bring Down the Next Digit:

Subtract 8 from 9 to get 1. Then bring down the next digit of 96, which is 6, to make 16.

4. Divide the New Number:

Determine how many times 4 goes into 16.

4 × 4 = 16

so 4 goes into 16 four times.

5. Write 4 above the division bracket next to 2.Multiply 4 by 4 and write the result (16) below 16.

6. Subtract Again:

Subtract 16 from 16 to get 0. There is no remainder.

The result is 24.

Place Value Chart Method:

Steps :

1. Set Up the Problem:

96 is represented as 9 tens and 6 ones.
4 is the divisor.

2. Represent 96 with Discs:

Place discs in the chart to represent the number 96:

9 discs in the tens place (representing 90).
6 discs in the ones place.

The chart looks like this:

3. Create 4 Groups as 4 is the Divisor:

Draw 4 groups for dividing the discs equally.

4. Distribute Tens and Ones discs in the place value chart

Place 9 tens and 6 ones discs in each group until the discs are evenly distributed.

5. Move the Extra Tens Disc to Ones

Bring the extra 1 ten to the ones column. Since 1 ten = 10 ones, add 10 ones to the existing 2 ones.

Now, distribute the 12 ones equally into the 4 groups in Ones place. Start by putting 1 one in each group, and repeat until all 12 ones are distributed equally.

We divided the 16 ones among the 4 groups.
Each group received 4 ones (16 ÷ 4 = 4 ones per group).
There is no remainder. All the ones were divided equally.

Since 4 groups of 24 (2 tens and 4 ones in each group) make up the total of 96, our answer is correct.

Thus, 96 ÷ 4 = 24

We know our answer is correct because 24 × 4 = 96.

Area Model

Steps:

Step 1: Break Down the Dividend (96) into Simpler Numbers

We will break 96 into smaller parts that are easy to divide by 4.
Let's split 96 into two numbers: 80 and 16.
96 = 80 + 16.
These numbers are easier to divide by 4.

Step 2: Apply Division to Each Part (Associative Property)

Now divide each part of the number bond by 4 separately:

80 ÷ 4 = 20

16 ÷ 4 = 4

Step 3: Add the Results Together

Combine the results of the two simpler divisions:

20+4=24

Conclusion:

So, 96 ÷ 4 = 24

This approach simplifies division by breaking down the larger number into smaller, more manageable parts, and then using the associative property to divide them individually before adding the results together.

Terminology

• Associative property E.g., 96 = 3 × (4 × 8) = (3 × 4) × 8.

• Composite number Positive integer having three or more whole number factors.

• Distributive property E.g., 64 × 27 = (60 × 20) + (60 × 7) + (4 × 20) + (4 × 7) .

• Divisible.

• Divisor The number by which another number is divided.

• Formula A mathematical rule expressed as an equation with numbers and/or variables.

• Long division Process of dividing a large dividend using several recorded steps.

• Partial product E.g., 24 × 6 = (20 × 6) + (4 × 6) = 120 + 24.

• Prime number Positive integer greater than 1 having whole number factors of only 1 and itself.

• Remainder The number left over when one integer is divided by another.

Tools and Representations