- Multiply and Divide Whole Numbers
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- Multiplication as Repeated Addition
- Single Digit Multiplication
- Multiplication by 10, 100, and 1000
- Multiplication Without Carry
- Multiplication With Carry
- Multiplication With Trailing Zeros: Problem Type 1
- Multiplication With Trailing Zeros: Problem Type 2
- Multiplication of 2-digit Numbers With 2-digit Numbers
- Multiplication of a Single Digit Number With Large Numbers
- Multiplication of Large Numbers
- Multiples Problem Type 1
- Multiples Problem Type 2
- Division Facts
- Fact Families for Multiplication and Division
- Multiplication or Division of Whole Numbers (Word problems)
- Multiplication and Addition or Subtraction of Whole numbers (Word problems)
- Unit Rates and Ratios of Whole Numbers (Word problems)
- Division Without Carry
- Division With Carry
- Division Involving Zero
- Whole Number Division: 2-digit by 2-digit, No Remainder
- Whole Number Division: 3-digit by 2-digit, No Remainder
- Division With Trailing Zeros: Problem Type 1
- Division With Trailing Zeros: Problem Type 2
- Quotient and Remainder: Problem type 1
- Quotient and Remainder: Problem type 2
- Quotient and Remainder (Word problems)

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Let us consider an example to understand multiplication as a repeated addition operation. Suppose there are some toy cars divided into 6 groups each having 4 toy cars. The total number of toy cars can be found by adding 4 repeatedly 6 times as shown below.

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

The same result is also obtained by multiplication operation. Since 4 is being repeatedly added take 4 and multiply it with the number of groups which is 6.

So 4 × 6 = 24 (Read as 4 times 6 equals 24)

Here 4 and 6 care called as **factors** and the resulting number is called as **product**.

The way **multiplication** is related to **repeated addition** can be explained in this way, i.e. multiplying **a × b** is the same as adding **a** repeatedly **b** number of times.

For **example**, for objects arranged in 4 rows and 5 columns (20 in all);

5 + 5 + 5 + 5 and 4 × 5 represents the total number of objects.

When you multiply, you add equal groups together to find the total.

Write the number of keys that you see. Write it as both an addition and multiplication problem

2 + 2 + 2 + 2 =

4 × 2 =

**Step 1:**

We see 2 keys in each group. There are 4 groups.

To simplify, there are 4 groups of 2 keys or

2 + 2 + 2 + 2 = 8

**Step 2:**

This can also be written as a multiplication problem.

4 groups and each group has 2 keys so we can multiply

4 × 2 = 8

**Step 3:**

So,

2 + 2 + 2 + 2 = 8

4 × 2 = 8

Rewrite the following repeated addition as a multiplication sentence

2 + 2 + 2 + 2 +2 + 2 = 12

**Step 1:**

Here 2 is being repeatedly added, so first write a 2. Then we count the number of times it is being added. This is 6 times.

**Step 2:**

So the multiplication sentence would be

2 × 6 = 12

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