Multiplication

In multiplication we know how to multiply a one, two or three-digit number by another 1 or 2-digit number. We also know how to multiply a four-digit number by a 2-digit number. We also know the different methods of multiplication. Here, we shall make use of the methods and procedures learnt previously in multiplying larger numbers.

The numbers that are being multiplied are called factors.
The answer of a multiplication operation is called a product.

We will recall how to do multiplication of a number by a 1-digit number.


Examples on Expanded Notation Method:

1. Use expanded notation to multiply 64 and 8.

Expanded notation



4 × 8   =    32

60 × 8 = + 480
          =    512

Answer: 512

3 and 4 digit numbers can also be multiplied by a 1 digit number using the expanded notation method.

2. Use expanded notation to multiply 413 by 5.

Expanded Notation to Multiply





3 × 5    =        15
10 × 5  =        50
400 × 5 = + 2000
            =   2065

Answer: 2065

3. Use expanded notation to multiply 1246 by 3.

Expanded Notation to Multiply





6 × 3      =       18
40 × 3    =       50
200 × 3  =      600
1000 × 3 = + 3000
             =    3738

Answer: 3738

4. Use expanded notation to find the product of 1409 and 5.

Expanded Notation





9 × 5     =       45
0 × 5     =         0
400 × 5  =    2000
1000 × 5 = + 5000
             =    7045

Answer: 7045

Let us first revise the process of multiplication.

Consider the following:

1. Multiply 215 by 7

Solution:

(i) (Expanded notation method)

215 × 7 = (200 + 10 + 5) × 7          1 4 0 0
= 200 × 7 + 10 × 7 + 5 × 7        +        7 0
= 1400 + 70 + 35                      +        3 5
= 1505                                         1 5 0 5


(ii) (Column method)

Column Method

Product = 1505

(i) 5 ones × 7 = 35 = 3 tens + 5 ones

5 is written under one column, 3 ten is carried over

(ii) 1 ten × 7 = 7 tens,

7 tens + 3 tens = 10 tens = 1 H + 0 ten.

0 is written under ten-column,

1 hundred is carried over

(iii) 2 hundreds × 7 = 14 hundreds

14 hundreds + 1 hundred = 15H

15H = 1Th + 5H. 1 is written under

Th-column and 5H is placed under H-column

So, 215 × 7 = 1505


2. Multiply 6103 by 8

Solution:

(i) (Expanded notation method)

6103 × 8 = (6000 + 100 + 0 + 3) × 8          4 8 0 0 0
= 48000 + 800 + 0 + 24                        +        8 0 0
= 48824                                             +             0
                                                        +          2 4
                                                            4 8 8 2 4


(ii) (Column method)

Multiplication

Product = 4 8 8 2 4

(i) 3 one × 8 = 24 ones = 2 tens + 4 ones

4 is placed under ones, 2 tens is carried over

(ii) 0 ten × 8 = 0, 0 + 2 = 2 tens, placed
under ten's

(iii) 1H × 8 = 8 hundreds, 8H is placed under H

(iv) 6Th × 8 = 48 thousands. It is placed under
Th.

So, 6103 × 8 = 48824


3. Find the product of 2113 and 3 using column method.

Solution:

multiply 4 digit numbers by a 1 digit number

Answer: 8452

3 × 4 = 12 ONES
Regroup as 1 TEN 2 ONES.

Related Concept

Addition

Word Problems on Addition

Subtraction

Check for Subtraction and Addition

Word Problems Involving Addition and Subtraction

Estimating Sums and Differences

Find the Missing Digits

Multiplication

Multiply a Number by a 2-Digit Number

Multiplication of a Number by a 3-Digit Number

Multiply a Number

Estimating Products

Word Problems on Multiplication

Multiplication and Division

Terms Used in Division

Division of Two-Digit by a One-Digit Numbers

Division of Four-Digit by a One-Digit Numbers

Division by 10 and 100 and 1000

Dividing Numbers

Estimating the Quotient

Division by Two-Digit Numbers

Word Problems on Division


4th Grade Math Activities

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