**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 3 – Digit Numbers?**

Before we come to understand the multiplication of 3 – digit numbers, it is important to recall what is meant by 3 – 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 3 – digit numbers? 2 – digit numbers are numbers that consist of 3 digits, i..e the numbers that consist of digits at the one’s place, the ten’s place and the hundred’s place only. For example, the numbers 545 and 673 are both 3 – digit numbers.

Now, let us move to learn the multiplication of 3 – digit numbers. When it comes to the multiplication of 3 – 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 635 by 400

Solution

We will solve this step by step.

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

635 = 600 + 30 + 5

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

600 x 400 + 30 x 400 + 5 x 400 = 240000 + 12000 + 2000

Step 3 – Add the results obtained. We get,

240000 + 12000 + 2000= 54000

**Hence, 635 x 400 = 54000**

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 341 by 120**

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 341 as the multiplicand and 120 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 0. We get

Step 3 – Similalry, we multiply the number at the hundred’s place and the ten’s place of the multiplicand by 0. 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 – We wil now repeat the steps 4 and 5 above to multiply the digit at the hundred’s place of the multiplicandd with the corrspondding digit of the multiplier. We will get –

Step 7 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, 341 x 120 = 40920**

**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 304 by 125**

Solution

We will use the following steps to obtain our result.

Step 1 – First we write the multiplicand and the multiplier in columns. Here, the multiplicand is 304 and the multipliers is 125

Step 2- Multiply the one’s digit of the multiplicand by 5. 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 5. We get 5 x 0 = 0. Add 2 that was carried over to it to get 0 + 2 = 2 Now, write 2 in the ten’s column to get

Step 4 – Multiply the hundred’s digit of the multiplicand by 5. We get 3 x 5 = 15. Now, write 5 in the hundred’s column and 1 in the ten thousand’s column.

Step 5 Now, we need to place 0 at the ones’ place in the next line as a placeholder. We will get

Step 6 – Next, we repeat the above steps to multiply all digits of the multiplicand by the digit at the ten’s place of the multiplier. We will get

Step 7 Now, we need to place 0 at the ones’ and the ten’s place in the next line as a placeholder. We will get

Step 8 – Again, we repeat the above steps to multiply all digits of the multiplicand by the digit at the hundred’s place of the multiplier. We will get

Step 9 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, 304 x 125 = 38000**

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 –

- First, we write the multiplicand and the multiplier in columns.
- First, multiply the number at the one’s place of the multiplier with all the numbers of the multiplicand and write them horizontally.
- Make sure you write numbers from right to left and each number below the corresponding place value of the multiplicand.
- Now, move to the next line.
- Place a 0 at the one’s place of this line.
- 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.
- Again move to the next line.
- Place a 0 at the one’s as well as ten’s place of this line.
- 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.
- Continue in this manner by adding an extra zero in each line until you have reached the end of the multiplier
- Add the numbers vertically according to their place values.
- The number so obtained is your result.

An important point to note here is that to multiply a number by 10, put 0 to the right of the number.

For example, 435 x 10 = 4350

Similarly, to multiply a number by 100, put two zeros to the right of the number.

For example, 435 x 100 = 43500

And, to multiply a number by 1000, put three zeros to the right of the number.

For example, 435 x 1000 = 435000

Let us now see an example where we have 3 digits in the multiplicand.

**For example, Multiply 364 by 123**

**Solution**

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. Place a 0 at the one’s place of the next line

4. 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.

5. Place a 0 at the one’s as well as ten’s place of the next line.

6. 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.

7. There is no more number in the multiplicand. Now, add the numbers vertically according to their place values.

8. Hence the final product is 44,772. We can say that 364 x 123 = 44772

**Solved Examples**

**Example 1** In a school a fee of £ 345 is collected per student. If there are 240 students in the school, how much fee is collected by the school?

**Solution** We have been given that in a school a fee of £ 345 is collected per student. Also, there are 240 students in the school. We need to find out the total fee collected by the school from all students. Let us first summarise this information

Amount of fee collected by the school from each student = £ 345

Number of students in the school = 240

Total amount of fee collected by the school = ?

This can be calculated by multiplying the fee collected for each student by the number of students in the school. Therefore we have,

Total amount of fee collected by the school = (Amount of fee collected by the school from each student ) x (Number of students in the school ) …….. ( 1 )

Substituting the given information in the above equation, we get

Total amount of fee collected by the school = £ ( 345 x 240 )

Now,

**Hence, Total amount of fee collected by the school = £ 82800**

**Example 2** A hawker delivers 148 newspapers every day. How many newspapers will he deliver in a non-leap year?

**Solution** We have been given that a hawker delivers 148 newspapers every day. We need to find out the total number of newspapers that he will deliver in a non-leap year. Let us summarise the given information as

Number of newspapers delivered by the hawker in a day = 148

Number of newspapers that he will deliver in a non-leap year = ?

Now, we know that a non-leap year consists of 365 days. This means that we need to find out the total number of newspapers that the hawker will deliver in 365 days. Therefore,

Total number of days on which hawker delivers the newspapers = 365

Now, to find the total number of newspapers delivered by the hawker in 365 days we will have to multiply the Number of newspapers delivered by the hawker in a day by the total number of days in a year. So, we have,

Number of newspapers that he will deliver in a non-leap year = (Number of newspapers delivered by the hawker in a day ) x (total number of days in a year ) ……….. ( 1 )

Substituting the given values in the above equation, we have

Number of newspapers that he will deliver in a non-leap year = 148 x 365

Now,

**Hence, Number of newspapers that he will deliver in a non-leap year = 54020**

**Key Facts and Summary**

- Multiplication is defined as the process of finding out the product between two or more numbers.
- The number to be multiplied is called the multiplicand.
- The number with which we multiply is called the multiplier.
- The result obtained after multiplying the multiplier and the multiplicand is called the product.
- 3 – digit numbers are numbers that consist of 3 digits.
- 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.
- In the column method, we split the numbers into columns and multiply the numbers by the multiplicand one by one.
- 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.

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