# 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.

Let us recall the properties of multiplication:

I. The product does not change when the order of the numbers is changed.

5 × 4 = 4 × 5

6 × 3 = 3 × 6

II. The product does not change when the grouping of numbers is changed.

(6 × 7) × 4 = 6 × (7 × 4)

2 × (5 × 3) = (2 × 5) × 3

III. The product of a number and 1 is the number itself.

96 × 1 = 96

72 × 1 = 72

IV. The product of a number and 0 is 0.

754 × 0 = 0

316 × 0 = 0

For examples:

40 × 5 = 200

400 × 5 = 2000

4000 × 5 = 20000

Now, 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.

 4 × 8   =    32 60 × 8 = + 480           =    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.

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

3. Use expanded notation to multiply 1246 by 3.

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

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

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

 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)

 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:

 Answer: 8452 3 × 4 = 12 ONES Regroup as 1 TEN 2 ONES.

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