Multiplicative Reasoning and Fluency

Multiplication as Equal Groups

Sometimes we need to count many things quickly. If the things are in equal groups, we can use multiplication instead of counting one by one.

Equal groups means each group has the same number of items. When groups are equal, multiplication helps us find the total.

For example, if there are 3 bags and each bag has 4 apples, we can count by ones: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. But multiplication is faster.

We can say:

There are 3 groups of 4.

That is written as \(3 \times 4\).

This means:

$$3 \times 4 = 12$$

So, 3 groups of 4 equals 12.

When we multiply equal groups, we are finding the total number of items.

How to think about multiplication as equal groups

• The first number tells how many groups there are.
• The second number tells how many items are in each group.
• The answer tells how many items there are altogether.

So in \(5 \times 2\):

• 5 = number of groups
• 2 = number in each group
• 10 = total

We can also show multiplication with repeated addition. Repeated addition means adding the same number again and again.

For example:

$$4 \times 3 = 3 + 3 + 3 + 3 = 12$$

This means 4 groups of 3.

Important idea: the groups must be equal. If one group has 2 items and another has 5 items, those are not equal groups, so that is not a multiplication equal-groups model.

Let’s build the idea step by step.

If you see 2 plates with 5 cookies on each plate, you have 2 equal groups of 5.

You can write:

$$2 \times 5 = 10$$

You can also count by fives: 5, 10.

If you see 4 boxes with 2 crayons in each box, you have 4 equal groups of 2.

You can write:

$$4 \times 2 = 8$$

You can also count by twos: 2, 4, 6, 8.

Worked Example 1

There are 3 baskets. Each basket has 2 pears. How many pears are there in all?

Step 1: Find the number of groups.

There are 3 baskets, so there are 3 groups.

Step 2: Find how many are in each group.

Each basket has 2 pears.

Step 3: Write the multiplication sentence.

\(3 \times 2\)

Step 4: Find the total.

$$3 \times 2 = 6$$

Answer: There are 6 pears.

Worked Example 2

There are 5 jars. Each jar has 3 marbles. How many marbles are there in all?

Step 1: Number of groups = 5

Step 2: Number in each group = 3

Step 3: Write the multiplication sentence.

\(5 \times 3\)

Step 4: Use repeated addition or skip-counting.

$$3 + 3 + 3 + 3 + 3 = 15$$

So,

$$5 \times 3 = 15$$

Answer: There are 15 marbles.

Worked Example 3

A teacher puts 4 pencils on each table. There are 6 tables. How many pencils are there altogether?

Step 1: Number of groups = 6 tables

Step 2: Number in each group = 4 pencils

Step 3: Write the multiplication sentence.

\(6 \times 4\)

Step 4: Skip-count by 4 six times.

4, 8, 12, 16, 20, 24

So,

$$6 \times 4 = 24$$

Answer: There are 24 pencils.

Worked Example 4

Sam says, “\(3 \times 5\) means 3 groups of 5.” Is Sam correct? How many items are there in all?

Yes, Sam is correct.

\(3 \times 5\) means:

• 3 groups
• 5 items in each group

Add the groups:

$$5 + 5 + 5 = 15$$

So,

$$3 \times 5 = 15$$

Answer: Sam is correct, and there are 15 items in all.

How to spot equal groups in a word problem

Look for words and ideas like these:

• each
• every
• same number
• groups of
• bags, boxes, baskets, plates, jars, rows, tables

These clues often tell you that multiplication can help.

Example clues:

• 4 bags with 6 oranges in each bag
• 2 rows of 7 chairs
• 8 boxes with 3 toys in each box

Each of these has equal groups.

A helpful way to solve

1. Find how many groups there are.
2. Find how many items are in each group.
3. Write a multiplication sentence.
4. Find the total by skip-counting or repeated addition.

Be careful!

• Do not just count random objects. Make sure the groups are equal.
• The total is not one of the numbers you multiply. The total is the answer.
• If there are 4 groups of 3, write \(4 \times 3\), not just 4 + 3.

Try thinking about these:

• 7 groups of 2 means \(7 \times 2\)
• 2 groups of 7 means \(2 \times 7\)

Both are multiplication sentences about equal groups. They may describe different group stories, but both can help find a total.

Summary

Multiplication helps us count equal groups quickly. The first number tells how many groups there are, and the second number tells how many items are in each group.

You can solve multiplication equal-group problems by using repeated addition or skip-counting. When you see the same number in each group, multiplication is a smart way to find the total.

Multiplication as Repeated Addition

Sometimes in math, we need to add the same number again and again. Doing that with a long addition sentence can take time.

Multiplication gives us a faster way to show repeated addition. It helps us count equal groups more easily.

For example, if you have 4 groups with 5 stars in each group, you could add:

\(5 + 5 + 5 + 5 = 20\)

We can write that with multiplication as:

\(4 \times 5 = 20\)

This means 4 groups of 5.

What multiplication means

In a multiplication sentence like \(3 \times 4\), the first number tells how many groups there are. The second number tells how many are in each group.

So \(3 \times 4\) means 3 groups of 4.

The repeated addition sentence is:

\(4 + 4 + 4 = 12\)

So,

$$3 \times 4 = 4 + 4 + 4 = 12$$

How to turn multiplication into repeated addition

1. Look at the first number. That tells how many groups to make.

2. Look at the second number. That tells how many are in each group.

3. Write the second number again and again for the number of groups.

4. Add to find the total.

Example: \(2 \times 6\)

• 2 groups

• 6 in each group

• Repeated addition: \(6 + 6\)

• Total: \(12\)

So, \(2 \times 6 = 6 + 6 = 12\).

Equal groups are important

We use multiplication when the groups are equal. That means each group has the same number.

If one group has 3 apples and another group has 5 apples, that is not multiplication as repeated addition, because the groups are not equal.

If 3 baskets each have 5 apples, that is multiplication, because the groups are equal.

Worked Example 1

Write \(3 \times 2\) as repeated addition and find the answer.

Step 1: The first number is 3, so there are 3 groups.

Step 2: The second number is 2, so each group has 2.

Step 3: Write repeated addition.

\(2 + 2 + 2\)

Step 4: Add.

$$2 + 2 + 2 = 6$$

So, \(3 \times 2 = 6\).

Worked Example 2

Write \(5 \times 3\) as repeated addition and find the answer.

There are 5 groups of 3, so we add 3 five times:

$$3 + 3 + 3 + 3 + 3 = 15$$

So, \(5 \times 3 = 15\).

Worked Example 3

There are 4 bags. Each bag has 6 marbles. How many marbles are there in all?

This means 4 equal groups of 6.

Multiplication sentence:

\(4 \times 6\)

Repeated addition sentence:

$$6 + 6 + 6 + 6 = 24$$

So there are 24 marbles in all.

Worked Example 4

Write the repeated addition for \(6 \times 4\), then solve.

\(6 \times 4\) means 6 groups of 4.

So we write 4 six times:

$$4 + 4 + 4 + 4 + 4 + 4 = 24$$

So, \(6 \times 4 = 24\).

A helpful way to think about it

You can say a multiplication fact as groups of.

• \(2 \times 5\) = 2 groups of 5 = \(5 + 5\)

• \(4 \times 3\) = 4 groups of 3 = \(3 + 3 + 3 + 3\)

• \(1 \times 7\) = 1 group of 7 = \(7\)

Be careful

• Do not add different numbers. Repeated addition uses the same number each time.

• Count the number of groups carefully.

• Make sure each group has the same amount.

Let’s compare

Look at \(4 \times 5\).

It means 4 groups of 5, so the repeated addition is:

\(5 + 5 + 5 + 5\)

That equals 20.

$$4 \times 5 = 5 + 5 + 5 + 5 = 20$$

This is much faster than thinking of 20 one by one. Multiplication helps us add equal groups quickly.

Summary

Multiplication is a quick way to show repeated addition.

In \(a \times b\), the first number tells how many groups there are, and the second number tells how many are in each group.

To solve, write the second number again and again for the number of groups, then add.

When groups are equal, repeated addition can help you understand and solve multiplication problems.

Arrays and the Area Model

Multiplication helps us count things faster when they are arranged in equal groups. One helpful way to see multiplication is with arrays.

An array is a set of objects arranged in rows and columns. Rows go across. Columns go up and down.

Arrays help us see multiplication clearly. They also help us understand the area model, which shows multiplication using a rectangle.

Let’s learn how arrays and the area model work.

1. What is an array?

If objects are lined up in equal rows, they make an array. Each row has the same number of objects.

For example, imagine 3 rows of 4 stars:

★ ★ ★ ★
★ ★ ★ ★
★ ★ ★ ★

This array has:

• 3 rows
• 4 columns

We can write this as:

\(3 \times 4 = 12\)

That means 3 rows of 4 make 12 in all.

We can also think about the same array as 4 columns of 3:

\(4 \times 3 = 12\)

The total stays the same. So arrays show us that when we turn the rows and columns, the product does not change.

2. Rows and columns

It is important to know the difference between rows and columns.

• Rows go across.
• Columns go up and down.

Look at this array:

● ● ●
● ● ●
● ● ●
● ● ●

This array has 4 rows and 3 columns.

So we can write:

\(4 \times 3 = 12\)

We can count by rows or by columns. Either way, there are 12 objects.

3. Arrays show equal groups

An array is another way to show equal groups. Instead of drawing circles with groups inside, we line the objects up in rows and columns.

For example, 2 rows of 5 dots means 2 equal groups of 5.

That is:

\(2 \times 5 = 10\)

Arrays help us count quickly because the objects are organized.

4. What is the area model?

The area model uses a rectangle to show multiplication. We split the rectangle into rows and columns, just like an array.

If a rectangle has 3 rows and 4 columns of square units, it matches the multiplication:

\(3 \times 4 = 12\)

We can show it like this:

□ □ □ □
□ □ □ □
□ □ □ □

There are 12 square units inside the rectangle. That is the area model.

The area model helps us see that multiplication can describe the space inside a rectangle made of equal squares.

5. Arrays and the area model are connected

An array and the area model are very similar. Both use rows and columns. Both help us multiply.

• An array can use objects like dots, stars, or counters.
• An area model uses a rectangle filled with square units.

Both can show the same multiplication sentence.

For example:

\(5 \times 2 = 10\)

This can mean:

• 5 rows of 2 objects in an array, or
• a rectangle with 5 rows and 2 columns of square units in an area model.

6. Breaking apart arrays

Sometimes a bigger array can be broken into smaller parts. This helps make multiplication easier.

For example, suppose we want to find:

\(4 \times 6\)

We can break 6 into 3 and 3.

Then we think:

$$4 \times 6 = 4 \times 3 + 4 \times 3$$

Now solve each smaller part:

\(4 \times 3 = 12\)

\(4 \times 3 = 12\)

Add them together:

\(12 + 12 = 24\)

So:

$$4 \times 6 = 24$$

This works with the area model too. We can split one big rectangle into two smaller rectangles.

Worked Example 1

Find the total in an array with 3 rows of 2.

Step 1: Write the multiplication sentence.

\(3 \times 2\)

Step 2: Count by rows.

2, 4, 6

Step 3: Write the answer.

$$3 \times 2 = 6$$

So an array with 3 rows of 2 has 6 objects.

Worked Example 2

A rectangle has 2 rows and 7 columns of square units. What is the total number of square units?

Step 1: Write the multiplication sentence.

\(2 \times 7\)

Step 2: Multiply or skip count.

7, 14

Step 3: Write the answer.

$$2 \times 7 = 14$$

The rectangle has 14 square units.

Worked Example 3

Look at an array with 5 rows of 4. How many objects are there?

Step 1: Write the multiplication sentence.

\(5 \times 4\)

Step 2: Skip count by 4 five times.

4, 8, 12, 16, 20

Step 3: Write the answer.

$$5 \times 4 = 20$$

There are 20 objects in the array.

Worked Example 4

Use breaking apart to solve \(3 \times 8\).

Step 1: Break 8 into 5 and 3.

Step 2: Write two smaller multiplication facts.

$$3 \times 8 = 3 \times 5 + 3 \times 3$$

Step 3: Solve each part.

\(3 \times 5 = 15\)

\(3 \times 3 = 9\)

Step 4: Add the parts.

\(15 + 9 = 24\)

So:

$$3 \times 8 = 24$$

Helpful tips

• Count rows across and columns up and down.
• Make sure each row has the same number of objects.
• Use skip counting to find the total.
• If a multiplication fact feels hard, break the array into smaller parts.
• An array of objects and a rectangle of square units can show the same multiplication.

Let’s practice thinking

• If there are 4 rows of 5, think: \(4 \times 5\).
• If there are 6 columns and 2 rows, think: \(2 \times 6\).
• If a rectangle is split into smaller rectangles, add the parts to find the whole.

Summary

Arrays are arrangements of objects in equal rows and columns. They help us see multiplication as equal groups.

The area model uses a rectangle made of square units to show the same idea. Both arrays and area models help us count, multiply, and break apart bigger problems into smaller ones.

When you see rows and columns, you can think multiplication.

Commutative Property of Multiplication

Today we will learn an important multiplication idea: changing the order of the factors does not change the total.

This is called the Commutative Property of Multiplication. That is a big name, but the idea is simple:

$$4 \times 3 = 3 \times 4$$

Both multiplication sentences have the same total, even though the numbers switch places.

We can understand this by thinking about equal groups and arrays.

Equal groups means the same number in each group. For example, \(4 \times 3\) means 4 groups of 3.

We can also show multiplication with an array. An array is a set of objects arranged in rows and columns.

For \(4 \times 3\), we can think of 4 rows with 3 in each row:

$$ \begin{array}{ccc} \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet \end{array} $$

This array has 4 rows and 3 columns. There are 12 dots altogether.

Now imagine turning, or rotating, the array 90 degrees. The rows and columns switch jobs.

After the turn, we see 3 rows with 4 in each row:

$$ \begin{array}{cccc} \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet \end{array} $$

Now the multiplication sentence is \(3 \times 4\). But the total number of dots is still 12.

So we can say:

$$4 \times 3 = 3 \times 4 = 12$$

The array changed its direction, but it did not change how many objects there were.

This is why the commutative property works. The factors can switch places, and the product stays the same.

Main idea:

• In multiplication, the numbers being multiplied are called factors.
• The answer is called the product.
• When we switch the factors, the product stays the same.

In math, we can write the rule like this:

$$a \times b = b \times a$$

This just means any two factors can switch places.

Why arrays help

Arrays make multiplication easy to see. One way shows rows. Another way shows columns. But both ways count the same total objects.

If an array has 2 rows and 5 columns, it has 10 objects. If you turn it, it becomes 5 rows and 2 columns. It still has 10 objects.

So:

$$2 \times 5 = 5 \times 2$$

Worked Example 1

Show that \(2 \times 4\) and \(4 \times 2\) are equal.

Start with 2 rows of 4:

$$ \begin{array}{cccc} \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet \end{array} $$

Count the dots: there are 8.

Rotate the array. Now we see 4 rows of 2:

$$ \begin{array}{cc} \bullet & \bullet \\ \bullet & \bullet \\ \bullet & \bullet \\ \bullet & \bullet \end{array} $$

Count again: there are still 8.

So:

$$2 \times 4 = 4 \times 2 = 8$$

Worked Example 2

Show that \(5 \times 3\) and \(3 \times 5\) are equal.

\(5 \times 3\) means 5 groups of 3. That is:

$$3 + 3 + 3 + 3 + 3 = 15$$

\(3 \times 5\) means 3 groups of 5. That is:

$$5 + 5 + 5 = 15$$

Both have the same total, so:

$$5 \times 3 = 3 \times 5$$

Worked Example 3

A student says, “\(6 \times 2\) is not the same as \(2 \times 6\) because the numbers are in a different order.” Is the student correct?

No, the student is not correct.

Let’s check:

$$6 \times 2 = 12$$

and

$$2 \times 6 = 12$$

They are equal. The order changed, but the product stayed the same.

You can picture 6 rows of 2 or 2 rows of 6. When the array is turned, the total number of objects stays 12.

Worked Example 4

Complete the sentence:

$$7 \times 4 = \square \times 7$$

Use the commutative property. Switch the factors.

$$7 \times 4 = 4 \times 7$$

So the missing number is 4.

Things to remember

• Multiplication can be shown with equal groups.
• Multiplication can be shown with arrays.
• Turning an array 90 degrees changes rows into columns.
• The total number of objects stays the same.
• So the order of factors does not change the product.

Helpful math facts

• \(1 \times 8 = 8 \times 1 = 8\)
• \(3 \times 6 = 6 \times 3 = 18\)
• \(4 \times 5 = 5 \times 4 = 20\)
• \(9 \times 2 = 2 \times 9 = 18\)

Learning one fact can help you know another fact. If you know \(3 \times 7 = 21\), then you also know \(7 \times 3 = 21\).

Brief Summary

The Commutative Property of Multiplication means you can switch the order of the factors, and the product stays the same.

Arrays help prove this. When you rotate an array 90 degrees, the rows and columns switch, but the total number of objects does not change.

That is why multiplication facts like \(4 \times 3\) and \(3 \times 4\) are equal.

Zero and Identity Properties help us multiply faster and understand what multiplication means.

In this lesson, we will learn two important rules:

• Identity Property of Multiplication: multiplying by 1 keeps the number the same.
• Zero Property of Multiplication: multiplying by 0 always gives 0.

These rules are very useful when solving multiplication problems.

First, let’s remember what multiplication means.

Multiplication means equal groups. For example, \(3 \times 4\) means 3 groups of 4.

If we draw it, it looks like this:

$$ 3 \times 4 = 4 + 4 + 4 = 12 $$

Now let’s use that idea to understand the two properties.

1. Identity Property of Multiplication

When we multiply any number by 1, the answer stays the same.

$$ 7 \times 1 = 7 $$ $$ 1 \times 7 = 7 $$

This is called the Identity Property because the number keeps its identity. It does not change.

Why does this happen?

Think about \(5 \times 1\). This means 5 groups of 1.

$$ 5 \times 1 = 1 + 1 + 1 + 1 + 1 = 5 $$

Now think about \(1 \times 5\). This means 1 group of 5.

$$ 1 \times 5 = 5 $$

In both cases, the total is still 5.

Here are more examples of the Identity Property:

• \(9 \times 1 = 9\)
• \(1 \times 9 = 9\)
• \(12 \times 1 = 12\)
• \(1 \times 12 = 12\)

Easy way to remember: multiplying by 1 means the number stays the same.

2. Zero Property of Multiplication

When we multiply any number by 0, the answer is always 0.

$$ 8 \times 0 = 0 $$ $$ 0 \times 8 = 0 $$

This is called the Zero Property of multiplication.

Why does this happen?

Think about \(4 \times 0\). This means 4 groups of 0.

Each group has nothing in it, so the total is nothing.

$$ 4 \times 0 = 0 + 0 + 0 + 0 = 0 $$

Now think about \(0 \times 4\). This means 0 groups of 4.

If there are no groups at all, there is still nothing to count.

$$ 0 \times 4 = 0 $$

Here are more examples of the Zero Property:

• \(6 \times 0 = 0\)
• \(0 \times 6 = 0\)
• \(15 \times 0 = 0\)
• \(0 \times 15 = 0\)

Easy way to remember: multiplying by 0 always gives 0.

Let’s compare the two properties.

• Multiply by 1 → the number stays the same.
• Multiply by 0 → the answer is always 0.

Look at these side by side:

$$ 6 \times 1 = 6 $$ $$ 6 \times 0 = 0 $$

The factor 6 is the same in both problems, but multiplying by 1 and multiplying by 0 give very different answers.

Worked Example 1

Solve: \(9 \times 1\)

We are multiplying by 1, so we use the Identity Property.

$$ 9 \times 1 = 9 $$

Answer: 9

Worked Example 2

Solve: \(7 \times 0\)

We are multiplying by 0, so we use the Zero Property.

$$ 7 \times 0 = 0 $$

Answer: 0

Worked Example 3

Solve: \(1 \times 12\)

This means 1 group of 12. One group of 12 is just 12.

$$ 1 \times 12 = 12 $$

Answer: 12

Worked Example 4

Solve: \(0 \times 10\)

This means 0 groups of 10. No groups means nothing to count.

$$ 0 \times 10 = 0 $$

Answer: 0

How to tell which property to use

1. Look at the multiplication problem.
2. If one factor is 1, use the Identity Property.
3. If one factor is 0, use the Zero Property.

Examples:

• \(13 \times 1\) → Identity Property → answer is \(13\)
• \(13 \times 0\) → Zero Property → answer is \(0\)
• \(1 \times 3\) → Identity Property → answer is \(3\)
• \(0 \times 3\) → Zero Property → answer is \(0\)

Be careful!

Sometimes students mix up the two properties.

• \(4 \times 1\) is not 0. It is \(4\).
• \(4 \times 0\) is not 4. It is \(0\).

A good memory trick is:

• 1 keeps the number alive.
• 0 makes the total zero.

Let’s practice thinking about groups.

• \(3 \times 1\): 3 groups of 1 = 3
• \(1 \times 3\): 1 group of 3 = 3
• \(3 \times 0\): 3 groups of 0 = 0
• \(0 \times 3\): 0 groups of 3 = 0

Summary

• The Identity Property says multiplying by 1 keeps the number the same.
• The Zero Property says multiplying by 0 always gives 0.
• These rules work no matter which side the 1 or 0 is on.

So remember:

$$ a \times 1 = a $$ $$ 1 \times a = a $$ $$ a \times 0 = 0 $$ $$ 0 \times a = 0 $$

When you see a multiplication problem with 1 or 0, you can solve it quickly by using these properties.

Multiplicative Comparison means comparing two amounts by telling how many times bigger one amount is than another amount.

For example, if Mia has 3 stickers and Jay has 9 stickers, we can say: Jay has 3 times as many stickers as Mia.

This is different from saying 6 more stickers. In 3rd Grade, it is very important to notice the difference between adding and multiplying when we compare amounts.

Additive comparison tells how many more or fewer.

• Example: 9 is 6 more than 3.

Multiplicative comparison tells how many times as many or as many times bigger.

• Example: 9 is 3 times as many as 3.

Both comparisons can be true for the same numbers, but they mean different things.

Let’s look at the numbers 4 and 12.

• 12 is 8 more than 4.
• 12 is 3 times as many as 4.

When you see words like times as many, times as much, or as many times, you should think about multiplication.

When you see words like more than or fewer than, you should think about addition or subtraction.

A helpful way to think about multiplicative comparison:

If one group has 3 objects, and another group has 4 times as many, then we need 4 groups of 3.

That means:

$$4 \times 3 = 12$$

So, 12 is 4 times as many as 3.

Use equal groups to help.

If Ben has 2 toy cars and Ava has 5 times as many, picture 5 equal groups of 2.

That is:

$$5 \times 2 = 10$$

Ava has 10 toy cars.

Worked Example 1

Lena has 4 crayons. Max has 3 times as many crayons as Lena. How many crayons does Max have?

Step 1: Find the smaller amount: 4.

Step 2: Multiply by 3 because Max has 3 times as many.

$$3 \times 4 = 12$$

Answer: Max has 12 crayons.

Worked Example 2

Noah has 5 shells. Emma has 15 shells. How can we compare Emma’s shells to Noah’s shells?

We ask: how many times as many is 15 as 5?

$$15 \div 5 = 3$$

So, Emma has 3 times as many shells as Noah.

We can also make an additive comparison:

$$15 - 5 = 10$$

Emma has 10 more shells than Noah.

Notice how the two comparison sentences are different:

• Emma has 10 more shells than Noah.
• Emma has 3 times as many shells as Noah.

Worked Example 3

A plant is 2 inches tall. Another plant is 8 inches tall. Which comparison is true?

• 8 is 6 more than 2.
• 8 is 4 times as many as 2.

Let’s check both:

$$8 - 2 = 6$$

So 8 is 6 more than 2.

$$4 \times 2 = 8$$

So 8 is also 4 times as many as 2.

Both are true, but one is additive and one is multiplicative.

Worked Example 4

Sam has 6 marbles. Eli has 18 marbles.

Is it correct to say, “Eli has 12 times as many marbles as Sam”?

Let’s check.

If Eli had 12 times as many, we would have:

$$12 \times 6 = 72$$

But Eli has 18 marbles, not 72.

So that sentence is not correct.

Now let’s find the true multiplicative comparison:

$$18 \div 6 = 3$$

Eli has 3 times as many marbles as Sam.

How to solve multiplicative comparison problems

1. Find the amount for one person or one group.
2. Look for the words times as many or times as much.
3. If you know the smaller amount and the number of times, multiply.
4. If you know both amounts and need to find how many times as many, divide.

Watch out for tricky words.

• 3 more than means add 3.
• 3 times as many means multiply by 3.

Example with 7:

• 3 more than 7 is $$7 + 3 = 10$$
• 3 times as many as 7 is $$3 \times 7 = 21$$

Those answers are very different, so reading carefully matters.

Try thinking with groups.

If a book has 4 pages in one section, and another section has 3 times as many pages, imagine 3 equal groups of 4.

$$3 \times 4 = 12$$

The other section has 12 pages.

Quick check ideas

• If the problem says times as many, your answer should usually be much bigger than the first amount.
• If you multiply, think about equal groups.
• If you divide, ask: “How many groups of the smaller number make the bigger number?”

Summary

Multiplicative comparison tells how many times as many one amount is compared to another amount.

Additive comparison tells how many more or fewer.

To solve multiplicative comparison problems, use multiplication or division and pay close attention to the words in the problem.

Deriving Unknown Facts via the Distributive Property

Sometimes a multiplication fact feels hard to remember. The good news is that you can break apart one factor into easier parts and then add the smaller products. This is called using the distributive property.

For example, if you want to solve \(7 \times 6\), you can think of \(7\) as \(5 + 2\). Then you multiply both parts by \(6\):

$$ 7 \times 6 = (5 \times 6) + (2 \times 6) $$

Now the harder fact becomes two easier facts.

This strategy helps you use facts you already know to find facts you do not know yet. It is like turning one big job into two smaller jobs.

What does “break apart” mean?

To break apart a factor means to split a number into two smaller numbers. The two smaller numbers must still make the original number.

• \(7 = 5 + 2\)
• \(8 = 5 + 3\)
• \(6 = 4 + 2\)
• \(9 = 5 + 4\)

Then you multiply each part by the other factor.

So instead of doing one unknown fact, you do two known facts and add them together.

The big idea

If one factor is broken into two parts, multiplication can be shared across both parts:

$$ (a+b) \times c = (a \times c) + (b \times c) $$

You do not need to memorize the rule with letters. Just remember this:

Break apart, multiply each part, then add.

Why this works

Imagine \(7 \times 6\) as 7 groups of 6. If you split the 7 groups into 5 groups and 2 groups, you still have the same total number of groups.

So:

$$ 7 \times 6 = 5 \times 6 + 2 \times 6 $$

You did not change the total. You only changed how you looked at it.

How to use this strategy

1. Look at the multiplication fact.
2. Choose one factor to break into two easier parts.
3. Multiply each part by the other factor.
4. Add the two products.

It often helps to break apart a factor into numbers you know well, like:

• \(5 + 1\)
• \(5 + 2\)
• \(5 + 3\)
• \(5 + 4\)
• \(2 + 2\)
• \(4 + 4\)

Worked Example 1

Find \(7 \times 6\).

Break apart \(7\) into \(5 + 2\):

$$ 7 \times 6 = (5 \times 6) + (2 \times 6) $$

Now solve the easier facts:

$$ 5 \times 6 = 30 $$ $$ 2 \times 6 = 12 $$

Add the products:

$$ 30 + 12 = 42 $$

So,

$$ 7 \times 6 = 42 $$

Worked Example 2

Find \(8 \times 4\).

Break apart \(8\) into \(5 + 3\):

$$ 8 \times 4 = (5 \times 4) + (3 \times 4) $$

Solve each part:

$$ 5 \times 4 = 20 $$ $$ 3 \times 4 = 12 $$

Add:

$$ 20 + 12 = 32 $$

So,

$$ 8 \times 4 = 32 $$

Worked Example 3

Find \(6 \times 7\).

This time, break apart \(6\) into \(4 + 2\):

$$ 6 \times 7 = (4 \times 7) + (2 \times 7) $$

Solve the known facts:

$$ 4 \times 7 = 28 $$ $$ 2 \times 7 = 14 $$

Add them:

$$ 28 + 14 = 42 $$

So,

$$ 6 \times 7 = 42 $$

Notice that \(6 \times 7\) and \(7 \times 6\) have the same answer. That is helpful too.

Worked Example 4

Find \(9 \times 3\).

Break apart \(9\) into \(5 + 4\):

$$ 9 \times 3 = (5 \times 3) + (4 \times 3) $$

Solve each part:

$$ 5 \times 3 = 15 $$ $$ 4 \times 3 = 12 $$

Add:

$$ 15 + 12 = 27 $$

So,

$$ 9 \times 3 = 27 $$

Using arrays to see it

An array is a set of rows and columns. Arrays can help you see how breaking apart works.

Think about \(7 \times 4\) as 7 rows of 4. You can split the 7 rows into 5 rows and 2 rows.

$$ 7 \times 4 = (5 \times 4) + (2 \times 4) $$

The whole array and the two smaller arrays have the same total number of squares.

Helpful choices for breaking apart numbers

When a fact feels tricky, choose a number split that gives you easy facts.

• For \(6\), try \(5+1\) or \(4+2\)
• For \(7\), try \(5+2\)
• For \(8\), try \(5+3\) or \(4+4\)
• For \(9\), try \(5+4\)

Many students like using \(5\) because facts with \(5\) are often easy to remember.

Let’s compare two ways

Suppose you need \(8 \times 6\).

You could break apart \(8\) as \(5+3\):

$$ 8 \times 6 = (5 \times 6) + (3 \times 6) = 30 + 18 = 48 $$

Or you could break apart \(6\) as \(5+1\):

$$ 8 \times 6 = (8 \times 5) + (8 \times 1) = 40 + 8 = 48 $$

Both ways work. Choose the way that feels easiest to you.

Common mistakes to watch out for

• Do not forget to multiply both parts. If \(7 = 5+2\), then \(7 \times 6\) becomes \((5 \times 6) + (2 \times 6)\), not \(5 \times 6 + 2\).
• Do not forget to add at the end. After finding the two smaller products, put them together.
• Make sure your parts equal the original number. If you break apart \(8\), your two parts must add to \(8\).

Try the thinking

If you wanted to solve \(7 \times 8\), you might think:

$$ 7 \times 8 = (5 \times 8) + (2 \times 8) $$ $$ 40 + 16 = 56 $$

That means \(7 \times 8 = 56\).

Remember

• Break one factor into two smaller parts.
• Multiply each part by the other factor.
• Add the two answers.

This is a smart way to use facts you know to find facts you are still learning.

Summary

The distributive property helps you solve harder multiplication facts by breaking one factor into easier parts. Then you multiply each part and add the products. For example, \(7 \times 6\) can be solved as \((5 \times 6) + (2 \times 6) = 30 + 12 = 42\).

Doubling and Halving Strategies help us solve multiplication problems in a smart, quick way. Instead of starting from the beginning every time, we can use a fact we already know and then double it or halve it to find a new fact.

This is a great strategy because many multiplication facts are connected. If you know one fact, you may be able to use it to figure out another one.

For example, if you know that \(2 \times 8 = 16\), then you can find \(4 \times 8\) by doubling 16. Since 16 doubled is 32, we get:

$$4 \times 8 = 32$$

Let’s learn how doubling and halving work.

Doubling means making a number 2 times as much. If you double 5, you get 10. If you double 12, you get 24.

Halving means splitting a number into 2 equal parts. If you halve 10, you get 5. If you halve 24, you get 12.

In multiplication, doubling and halving can help when one factor changes in a helpful way.

• If one factor doubles, the product doubles.
• If one factor halves, the product halves.

Look at this pattern with 6:

• \(2 \times 6 = 12\)
• \(4 \times 6 = 24\)
• \(8 \times 6 = 48\)

Each time the first factor doubles, the answer also doubles.

This strategy is helpful when you know easier facts like the 2s facts, 5s facts, or 10s facts. Then you can use them to solve harder facts.

How to use doubling:

1. Start with a multiplication fact you already know.
2. Find a new fact where one factor is doubled.
3. Double the product to get the new answer.

Worked Example 1

Find \(4 \times 7\).

Start with a fact you may know: \(2 \times 7 = 14\).

Since 4 is double 2, double the product 14:

$$14 + 14 = 28$$

So:

$$4 \times 7 = 28$$

Worked Example 2

Find \(8 \times 3\).

You might know \(4 \times 3 = 12\).

Since 8 is double 4, double 12:

$$12 + 12 = 24$$

So:

$$8 \times 3 = 24$$

How to use halving:

1. Look at a multiplication fact you need to solve.
2. See if one factor can be halved into an easier fact you know.
3. Halve the product of the harder fact’s partner fact, or solve the easier fact first and use it to build back up.

Halving helps us connect facts too. For example, if you know \(8 \times 5 = 40\), then you can find \(4 \times 5\) by halving 40.

Half of 40 is 20, so:

$$4 \times 5 = 20$$

Worked Example 3

Find \(6 \times 4\).

You may know \(3 \times 4 = 12\).

Since 6 is double 3, double 12:

$$12 + 12 = 24$$

So:

$$6 \times 4 = 24$$

Worked Example 4

Find \(5 \times 8\).

You may know \(10 \times 8 = 80\).

Since 5 is half of 10, take half of 80:

$$80 \div 2 = 40$$

So:

$$5 \times 8 = 40$$

We can also think about doubling and halving with equal groups.

If you have 2 groups of 9, that is:

$$2 \times 9 = 18$$

If you double the number of groups, you get 4 groups of 9. That doubles the total too:

$$4 \times 9 = 36$$

If you halve 4 groups of 9, you go back to 2 groups of 9, and the total is halved:

$$36 \div 2 = 18$$

Tips for using this strategy:

• Use facts you know well, like the 2s, 5s, and 10s.
• Ask yourself, “Can I double a smaller fact?”
• Ask yourself, “Can I halve a bigger fact?”
• Check that only one factor is doubling or halving when you use this strategy.

Be careful:

If both factors change, you cannot just double or halve once and expect the answer to work. Watch which number is changing.

For example:

• From \(2 \times 4 = 8\) to \(4 \times 4\), only one factor doubles, so the product doubles to 16.
• But from \(2 \times 4 = 8\) to \(4 \times 8\), both factors changed, so you need a different plan.

A good way to practice is to look for facts that are connected:

• \(2 \times 6 = 12\), so \(4 \times 6 = 24\)
• \(4 \times 6 = 24\), so \(8 \times 6 = 48\)
• \(10 \times 3 = 30\), so \(5 \times 3 = 15\)
• \(6 \times 7 = 42\), so \(3 \times 7 = 21\)

Summary

Doubling and halving strategies help you use multiplication facts you already know to find new ones. If one factor doubles, the product doubles. If one factor halves, the product halves.

These strategies make multiplication faster and easier. They help you see how facts are connected, and that is an important part of becoming strong at multiplication.

Multiplying by Multiples of 10

Sometimes we multiply by numbers like 10, 20, 30, or 40. These are called multiples of 10 because they are made from groups of 10.

When we multiply by a multiple of 10, we can use what we already know about basic multiplication facts. Then we use place value to find the full answer.

For example, 40 means 4 tens. So if we see \(3 \times 40\), we can think, “What is 3 groups of 4 tens?”

Big Idea: Multiply the one-digit numbers first, then think about tens.

Here is the pattern:

$$ 3 \times 40 = 3 \times 4\text{ tens} = 12\text{ tens} = 120 $$

This works because 40 is not just 4. It is 4 tens.

Step-by-step way to solve:

• Look at the multiple of 10.
• Find how many tens it has.
• Multiply the basic fact.
• Rename the answer in tens, then write the full number.

Let’s look at some multiples of 10:

• \(10 = 1\) ten
• \(20 = 2\) tens
• \(30 = 3\) tens
• \(40 = 4\) tens
• \(50 = 5\) tens
• \(60 = 6\) tens
• \(70 = 7\) tens
• \(80 = 8\) tens
• \(90 = 9\) tens

So when you multiply by one of these numbers, you are really multiplying by tens.

Worked Example 1

Solve \(2 \times 30\).

First, think of 30 as 3 tens.

Now multiply the basic fact: \(2 \times 3 = 6\).

That means \(2 \times 30 = 2 \times 3\text{ tens} = 6\text{ tens}\).

And 6 tens is 60.

$$ 2 \times 30 = 60 $$

Worked Example 2

Solve \(4 \times 20\).

20 is 2 tens.

Multiply the basic fact: \(4 \times 2 = 8\).

So we have 8 tens.

8 tens is 80.

$$ 4 \times 20 = 80 $$

Worked Example 3

Solve \(6 \times 40\).

40 is 4 tens.

Multiply the basic fact: \(6 \times 4 = 24\).

This means 24 tens.

24 tens is 240.

$$ 6 \times 40 = 240 $$

Worked Example 4

Solve \(9 \times 70\).

70 is 7 tens.

Multiply the basic fact: \(9 \times 7 = 63\).

So the answer is 63 tens.

63 tens is 630.

$$ 9 \times 70 = 630 $$

Another way to think about it

You may notice a shortcut. First multiply the non-zero digits, then place one zero at the end.

For example:

• \(3 \times 40\): first do \(3 \times 4 = 12\), then write 120
• \(5 \times 60\): first do \(5 \times 6 = 30\), then write 300

This works because the number already has one ten in it.

Be careful!

• Do not forget that 40 means 4 tens, not just 4.
• Do not stop at the basic fact. For example, in \(3 \times 40\), \(3 \times 4 = 12\), but the answer is not 12. It is 12 tens, which is 120.

Let’s compare:

• \(3 \times 4 = 12\)
• \(3 \times 40 = 120\)

The second answer is bigger because 40 is ten times as much as 4.

Try thinking with groups

If you have 5 groups of 20, that means 5 groups of 2 tens.

\(5 \times 2 = 10\), so that is 10 tens.

10 tens equals 100.

$$ 5 \times 20 = 100 $$

Helpful strategy

1. Circle the one-digit numbers you know.
2. Multiply those digits.
3. Remember the multiple of 10 means tens.
4. Write the answer with the correct zero.

Practice these on your own:

• \(3 \times 20\)
• \(7 \times 30\)
• \(8 \times 50\)
• \(4 \times 90\)

Answers:

• \(3 \times 20 = 60\)
• \(7 \times 30 = 210\)
• \(8 \times 50 = 400\)
• \(4 \times 90 = 360\)

Summary

Multiples of 10 are numbers like 10, 20, 30, and 40. To multiply by a multiple of 10, multiply the basic fact first, then think about tens. This helps you use place value to find the correct answer.

Fact Families and Missing Factors

In multiplication, numbers work together in special ways. Sometimes you know all the numbers, and sometimes one number is missing. In this lesson, you will learn how to use fact families to find a missing factor.

A factor is a number that is multiplied by another number. In the equation \(5 \times 7 = 35\), the factors are \(5\) and \(7\). The answer, \(35\), is called the product.

A fact family is a group of math facts that use the same numbers. Fact families help us see how multiplication and division are connected.

For example, the numbers \(4\), \(6\), and \(24\) make this fact family:

$$ 4 \times 6 = 24 $$ $$ 6 \times 4 = 24 $$ $$ 24 \div 4 = 6 $$ $$ 24 \div 6 = 4 $$

These four equations belong together because they use the same three numbers.

When a factor is missing, you can use what you know about multiplication facts or think about the matching division fact.

For example, in \(5 \times ? = 35\), you are asking, "What number times 5 equals 35?"

You can think:

• \(5, 10, 15, 20, 25, 30, 35\)
• That is \(7\) groups of \(5\).
• So the missing factor is \(7\).

You can also use division:

$$ 35 \div 5 = 7 $$

So:

$$ 5 \times 7 = 35 $$

This is why fact families are so helpful. They let you move between multiplication and division to find the missing number.

How to Find a Missing Factor

1. Look at the multiplication equation.
2. Ask, "What number is missing?"
3. Think of the related division fact, or use a multiplication fact you know.
4. Check your answer by multiplying.

Here is a simple way to think about it:

If

$$ a \times ? = c $$

then you can find the missing factor by thinking:

$$ c \div a = ? $$

You do not need to guess. You can use the relationship between the numbers.

Worked Example 1

Find the missing factor:

$$ 3 \times ? = 12 $$

Think: What number times \(3\) equals \(12\)?

We know:

$$ 3 \times 4 = 12 $$

So the missing factor is 4.

Check:

$$ 12 \div 3 = 4 $$

Worked Example 2

Find the missing factor:

$$ ? \times 8 = 40 $$

Think: How many groups of \(8\) make \(40\)?

Count by 8s:

• \(8\)
• \(16\)
• \(24\)
• \(32\)
• \(40\)

That is 5 groups of \(8\).

So:

$$ 5 \times 8 = 40 $$

The missing factor is 5.

Worked Example 3

Find the missing factor:

$$ 7 \times ? = 56 $$

Use the related division fact:

$$ 56 \div 7 = 8 $$

So:

$$ 7 \times 8 = 56 $$

The missing factor is 8.

Worked Example 4

The fact family uses \(9\), \(6\), and \(54\).

Write the fact family:

$$ 9 \times 6 = 54 $$ $$ 6 \times 9 = 54 $$ $$ 54 \div 9 = 6 $$ $$ 54 \div 6 = 9 $$

Now find the missing factor:

$$ 54 \div 6 = ? $$

The answer is 9.

This shows that multiplication and division facts in the same family help each other.

Tips to Remember

• A fact family uses the same three numbers.
• Multiplication and division are connected.
• A missing factor can be found with a related division fact.
• Always check by multiplying.

Let’s Practice Thinking

If you see \(4 \times ? = 28\), think:

$$ 28 \div 4 = 7 $$

So the missing factor is \(7\).

If you see \(? \times 3 = 18\), think:

$$ 18 \div 3 = 6 $$

So the missing factor is \(6\).

Summary

Fact families help you see how multiplication and division belong together. When a factor is missing, you can use a related division fact to find it. Instead of guessing, use what the numbers tell you. That is a smart math strategy.