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 |
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.
 |
3 × 5 = 15 |
Answer: 2065
3. Use expanded notation to multiply 1246 by 3.
 |
6 × 3 = 18 |
Answer: 3738
4. Use expanded notation to find the product of 1409 and 5.
 |
9 × 5 = 45 |
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)
 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 (iii) 1H × 8 = 8 hundreds, 8H is placed under H (iv) 6Th × 8 = 48 thousands. It is placed under |
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. |
Related Concept
● Addition
● Check
for Subtraction and Addition
● Word
Problems Involving Addition and Subtraction
● Estimating
Sums and Differences
● Multiply
a Number by a 2-Digit Number
● Multiplication
of a Number by a 3-Digit Number
● Word
Problems on Multiplication
● 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
● Division
by Two-Digit Numbers
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