Multiplication of Two Digit Numbers
We know that multiplication is one of the four basic operations in mathematics, the others being addition, subtraction, and division. We have studied addition and subtraction of numbers in earlier grades.
Multiplication is the repeated addition of the same number. When we add or combine groups of equal sizes successively, the product is called the multiplication product.
For example, when we add the number 3, 5 times repeatedly, we get,
3 + 3 + 3 + 3 + 3 = 15
As there are 5 equal groups of number 3, we write that as,
3 × 5 = 15
Multiplication of 2 Digit numbers
Multiplying two-digit numbers can be done easily using the partial product method.
What is multiplication using a partial product?
To multiply two numbers, we separate one of the numbers into parts and find the product with each. Later, we add them together to find the actual multiplication product. This is called the multiplication of numbers using partial products.
For example, we multiply 25 × 6, then
Steps to Multiply Two Digit Numbers
Step 1: Estimate the product using the nearest ‘tens‘ to verify the answer.
Step 2: Multiply any one of the numbers by the ‘ones’ digit of the second number.
Step 3: Multiply the number by the ‘tens’ digit of the number.
Step 4: Finally, add the two partial products to get the final product and verify it with the estimated product.
For example, we multiply 67 × 56 in the following steps:,
First, estimate the product as,
67 × 56
≈ 70 × 55
= 3850
Multiply 67 by 6 ‘ones’ or 6.
6 × 7 ‘ones’ = 42 ‘ones’.
Regroup as 4 ‘tens’ and 2 ‘ones’.
6 × 6 ‘tens’ = 36 ‘tens’.
Add the regrouped ‘tens’: 36 + 4 = 40
Then, multiply 67 by 50.
5 ‘tens’ × 7 = 35 ‘tens’.
Regroup as 3 ‘hundreds’ and 5 ‘tens’ or 50.
5 ‘tens’ × 60 = 300 ‘tens’ or 30 ‘hundreds’
Add the regrouped ‘hundreds’:
30 + 3 = 33.
Finally, add the partial product,
Align the partial products in the correct place values. Then add them to find the product.
As 3752 is close to the estimate 3850, the answer is reasonable.
Solved Double Digit Multiplication Examples
Example 1: Find 57 × 35
Solution:
Estimate: 60 × 35 = 2,100
Think: 57 is 5 ‘tens’ and 7 ‘ones’. 35 is 3 ‘tens’ and 5 ‘ones’.
Multiply 57 by 5 ‘ones’ or 5.
7 × 5 ‘ones’ = 35 ‘ones’. Regroup as 3 ‘tens’ and 5 ‘ones’.
5 × 5 ‘tens’ = 25 ‘tens’. Add the regrouped ‘tens’: 25 + 3 = 28
Multiply 57 by 3 ‘tens’ or 30,
3 ‘tens’ × 7 = 21 ‘tens’. Regroup as 2 ‘hundreds’ and 1 ‘tens’ or 10.
3 ‘tens’ × 50 = 150 ‘tens’ or 15 ‘hundreds’
Add the partial products.
Align the partial products in the correct place values. Then add them to find the product.
Since 1995 is close to the estimate 2100, the answer is reasonable.
Example 2:
Estimate and find the product.
41 × 32
Solution:
Estimate: 40 × 30 = 1200
Multiply 41 by 2 ‘ones’ or 2,
Multiply 41 by 3 ‘tens’, or 30
Write at the correct place value.
Add the two partial products.
The product is 1312. Since it is close to 1200, the answer is reasonable.
Example 3: Andrew eats 25 chocolates at a party. Each chocolate has 33 calories. How many calories does Andrew eat at the party?
Solution:
Andrew eats 25 chocolates. Each contains 33 calories. To find the total calories taken by Andrew, we multiply both the numbers.
To multiply, 25 × 33
Multiply 25 by 3 ‘ones’ or 3,
Multiply 25 by 3 ‘tens’ or 30.
Write at the correct place value.
Add the two partial products.
Hence, Andrew eats 825 calories at the party.
Example 4: Separate the problem into simpler parts and solve.
60 × 28
Solution:
60 × 28 = 60 × (20 + 8) [28 = 20 + 8]
= 60 × 20 + 60 × 8 [Use partial product]
= 1200 + 480 [Find each partial product]
= 1680 [Add each partial product]
Therefore, the product is 1680.
To multiply two numbers, we separate one of the numbers into parts of ‘tens’, ‘hundreds’, and so on. Then find the product with each part separately. Finally, add them together to find the multiplication product.
This is called the multiplication of numbers using partial products.
For example, we multiply 25 × 6, then
When two-digit numbers are multiplied through regrouping, the last digit of the second partial product will be 0. As the last digit is a multiplication product of ‘tens’ of numbers, it contains a 0 in the ‘ones’ place value.
It is a method of multiplication in which we separate a number into groups and multiply each separately. Later, we add the partial products to find the solution.
For example, we can use the long multiplication method to multiply 45 and 56 as,
