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# Multiplication of Two-Digit Numbers

## Introduction

Multiplication is one of the four basic operations in mathematics with the other three being addition, subtraction and division. Before we move to learn how to multiply 2 – digit numbers, let us recall what we mean by multiplication.

## How do we define Multiplication?

Multiplication is defined as the process of finding out the product between two or more numbers. The result thus obtained is called the product. Suppose you bought 6 pens on one day and 6 pens on the next day. Total pens you bought are now 2 times 6 or 6 + 6 = 12.

This can also be written as 2 x 6 = 12

Not the symbol used for multiplication. The symbol (x) is generally used to represent multiplication. Other common symbols that are used for multiplication are the asterisk (*) and dot (.)

### Symbol for Multiplication

Note the symbol used in the example above for multiplication. The symbol (x) is generally used to represent multiplication. Other common symbols that are used for multiplication are the asterisk (*) and dot (.)

Now, let us have a look at some important terms that are used when two numbers are multiplied.

### Important terms in the multiplication

Some important terms used in multiplication are –

Multiplicand – The number to be multiplied is called the multiplicand.

Multiplier – The number with which we multiply is called the multiplier.

Product – The result obtained after multiplying the multiplier and the multiplicand is called the product.

The relation between the multiplier, multiplicand and the product can be expressed as –

Multiplier × Multiplicand = Product

Let us understand this using an example.

Suppose we have two numbers 9 and 5. We wish to multiply 9 by 5.

So, we express it as 9 x 5 which gives us 45.

Therefore, 9 x 5 = 45

Here, 9 is the multiplicand, 5 is the multiplier and 45 is the product.

Now, that we have understood what we mean by multiplication and the terms associated with it, let us move to learn multiplication of 1 – digit numbers.

Now, let us understand how to perform multiplication when we have multi-digit numbers

## How to multiply 2 – Digit Numbers?

Before we come to understand the multiplication of 2 – digit numbers, it is important to recall what is meant by 2 – digit numbers?

Recall that every digit of a number has a place value. For instance, the number 5 is a single digit number where 5 is at the one’s place. Similarly, in number 27, the digit 2 is at the ten’s place while the digit 7 is at the one’s place. So, how do we define 2 – digit numbers? 2 – digit numbers are numbers that consist of 2 digits, i..e the numbers that consist of digits at the one’s place and the ten’s place only. For example, the numbers 55 and 67 are both 2 – digit numbers.

Now, let us move to learn the multiplication of 2 – digit numbers. When it comes to the multiplication of 2 – digit numbers, there are two methods of multiplying the numbers. These methods are the expanded notation method and the column method. Let us understand both the methods.

### Expanded Notation Method

In the expanded notation method we expand the multiplicand as per the place values and then multiply each number by the multiplier. We then sum up all the results obtained to get our final answer. Let us understand it through an example.

For example, Multiply 35 by 40

Solution

We will solve this step by step.

Step 1 – Write the number (multiplicand) in the expanded form. We get,

35 = 30 + 5

Step 2 – Multiply each number by the given number (multiplier) one by one. We get,

30 x 40 + 5 x 40 = 1200 + 200

Step 3 – Add the results obtained. We get,

1200 + 200 = 1400

Hence, 35 x 40 = 1400

This method, though simple, may not be suited for larger numbers. But it is used to understand the basic concepts of multiplication.

### Column Method

In this method, we split the numbers into columns and multiply the numbers by the multiplicand one by one. There are two scenarios when using this method.

Let us understand them one by one

#### Multiplication without Regrouping

This method comes into force when we have smaller numbers that do not involve carrying forward any numbers to the digit at the next place value. Let us understand it through an example.

For example, Multiply 21 by 32

Solution

We will use the following steps to obtain our result.

Step 1 – First we write the multiplicand and the multiplier in columns.  Here we have 21 as the multiplicand and 32 as the multiplier.

Step 2 – Now, we multiply the number at the one’s place of the multiplicand, i.e. 1 by the number at the one’s place of the multiplier, which in this case is 2. We get

Step 3 – Now, we multiply the number at the ten’s place of the multiplicand by 2. We get

Step 4 – Now, we need to place 0 at the ones’ place in the next line as a placeholder. We will get

Step 5 Since, we are complete with the multiplication of the multiplicand with the first digit of the multiplier, we perform the same steps as above for the multiplication of the multiplicand with the next number of the multiplier and then write the result in the line against the 0 that we placed as a placeholder in the previous step. We will get –

Step 6 Now that we have multiplied all the digits of the multiplier with the multiplicand, we will add the digits obtained in a vertical manner. We will get

The result thus obtained is our answer. Hence 21 x 32 = 672

#### Multiplication with Regrouping

In the above case, we have small multiplications that did not involve two-digit results at any step. But in the case of larger numbers, there will be a need to carry forward the number to the number at the next place value. This is called Multiplication with Regrouping. Let us understand it through an example.

For example, Multiply 25 by 34

Solution

We will use the following steps to obtain our result.

Step 1 – First we write the multiplicand and the multiplier in columns.

Step 2- Multiply the one’s digit of the multiplicand by 4. We have 4 x 5 = 20. Write 0 in the one’s column and carry over 2 to the ten’s column.

Step 3 – Multiply ten’s digit of the multiplicand by 4. We get 2 x 4 = 8. Add 2 that was carried over to it to get 8 + 2 = 10 Now, write 0 in the ten’s column and carry over 1 to the hundred’s column.

Step 4 Now, we need to place 0 at the ones’ place in the next line as a placeholder. We will get

Step 5 Since, we are complete with the multiplication of the multiplicand with the first digit of the multiplier, we perform the same steps as above for the multiplication of the multiplicand with the next number of the multiplier and then write the result in the line against the 0 that we placed as a placeholder in the previous step. We will get –

Step 6 Now that we have multiplied all the digits of the multiplier with the multiplicand, we will add the digits obtained in a vertical manner. We will get

The result thus obtained is our answer. Hence 25 x 34 = 850

The above steps can be generalised to define multiplication, which is commonly known as long multiplication. Let us define these steps.

## Long Multiplication

Long multiplication is similar to the column method except for the fact that here we multiply the larger numbers. This method is used when the multiplicand is greater than 9, i.e. the multiplicand is more than a one-digit number.  This method involves the following steps –

1. First, we write the multiplicand and the multiplier in columns.
2. First, multiply the number at the one’s place of the multiplier with all the numbers of the multiplicand and write them horizontally.
3. Make sure you write numbers from right to left and each number below the corresponding place value of the multiplicand.
4. Now, move to the next line.
5. Place a 0 at the one’s place of this line.
6. Now, look for the digit at the ten’s place of the multiplier. Multiply the number at the ten’s place of the multiplier with all the numbers of the multiplicand and write them horizontally in the line where you had marked 0.
7. Again move to the next line.
8. Place a 0 at the one’s as well as ten’s place of this line.
9. Now, look for the digit at the hundred’s place of the multiplier. Multiply the number at the hundred’s place of the multiplier with all the numbers of the multiplicand and write them horizontally in the line where you had marked the two zeros.
10. Continue in this manner by adding an extra zero in each line until you have reached the end of the multiplier
11. Add the numbers vertically according to their place values.
12. The number so obtained is your result.

Let us understand this by an example

For example, Multiply 32 by 13

Solution

1. First, we write the multiplicand and the multiplier in columns.
1. Then, multiply the number at the one’s place of the multiplier with all the numbers of the multiplicand and write them horizontally.
1. Place a 0 at the one’s place of the next line
1. Now, look for the digit at the ten’s place of the multiplier. Multiply the number at the ten’s place of the multiplier with all the numbers of the multiplicand and write them horizontally in the line where you had marked 0.
1. There is no more number in the multiplicand. Now, add the numbers vertically according to their place values.

The final answer is 416. Hence 32 x 13 = 416

Let us see another example where we 3 digits in the multiplicand.

For example, Multiply 53 by 25

Solution

1. First we write the multiplicand and the multiplier in columns

1. Then we multiply the number at the one’s place of the multiplier with all the numbers of the multiplicand and write them horizontally.
1. Place a 0 at the one’s place of the next line
1. Now, look for the digit at the ten’s place of the multiplier. Multiply the number at the ten’s place of the multiplier with all the numbers of the multiplicand and write them horizontally in the line where you had marked 0.
1. Now that we have multiplied all the digits of the multiplier with the multiplicand, we will add the digits obtained in a vertical manner. We will get

Hence, 53 x 25 = 1325

## Solved Examples

Example 1 There are 58 rows in a cinema hall and there are 25 seats in each row. How many persons can be accommodated in the hall?

Solution We have been given that there are 58 rows in a cinema hall and there are 25 seats in each row. We need to find the number of persons that can be accommodated in the hall. Let us summarise the information given to us.

Number of rows in cinema hall = 58

Number of seats in each row = 25

The number of persons that can be accommodated in the hall = ?

We can find the number of persons that can be accommodated in the hall by multiplying the number of rows with the number of seats in each row. This means that –

Number of persons that can be accommodated in the hall =  ( Number of rows in cinema hall ) x (Number of seats in each row ) …………………… ( 1 )

Substituting the given values in the above equation, we will have,

Number of persons that can be accommodated in the hall =  58 x 25

Now,

Hence, the number of persons that can be accommodated in the hall = 1450

Example 2 A restaurant purchased 81 boxes of ketchup packets. Each box contained 49 packets of ketchup. How many ketchup packets in total did the restaurant purchase?

Solution We have been given that a restaurant purchased 81 boxes of ketchup packets. Each box contained 49 packets of ketchup. We need to find out how many ketchup packets in total did the restaurant purchase. Let us summarise the information given to us.

Number of boxes of ketchup packets purchased by the restaurant = 81

Number of packets of ketchup in each box = 49

The total number of packets of ketchup purchased by the restaurant = ?

To find the value of the number of packets of ketchup purchased by the restaurant, we will have to multiply the number of boxes of ketchup packets purchased by the restaurant by the number of packets of ketchup in each box. This means that

Total number of packets of ketchup purchased by the restaurant = (Number of boxes of ketchup packets purchased by the restaurant ) x (Number of packets of ketchup in each box ) ………………………….. ( 1 )

Substituting the given values in the above equation, we will have,

Total number of packets of ketchup purchased by the restaurant = 81 x 49

Now,

Hence, the total number of packets of ketchup purchased by the restaurant = 3969.

Example 3 William bought 60 packages of paper tissues. There were 56 tissues in each package. How many tissues did William buy?

Solution We have been given that William bought 60 packages of paper tissues. There were 56 tissues in each package. We need to find the number of tissues bought by William. Let us summarise the information given to us.

Number of packages of paper tissues bought by William = 60

Number of tissues in each package = 56

The total number of tissues bought by William = ?

To find the total number of tissues bought by William, we will have to multiply the number of packages of paper tissues bought by William by the number of tissues in each package. This means that

Total number of tissues bought by William = (Number of packages of paper tissues bought by William ) x (Number of tissues in each package ) ………. ( 1)

Substituting the given values in the above equation, we will have,

Total number of tissues bought by William = 60 x 56

Now,

Hence, the total number of tissues bought by William = 3360

## Key Facts and Summary

1. Multiplication is defined as the process of finding out the product between two or more numbers.
2. The number to be multiplied is called the multiplicand.
3. The number with which we multiply is called the multiplier.
4. The result obtained after multiplying the multiplier and the multiplicand is called the product.
5. 2 – digit numbers are numbers that consist of two digits, one digit at the ten’s place and one digit at the one’s place.
6. In the expanded notation method we expand the multiplicand as per the place values and then multiply each number by the multiplier. We then sum up all the results obtained to get our final answer.
7. In the column method, we split the numbers into columns and multiply the numbers by the multiplicand one by one.