Tap to Read ➤

Anup Patwardhan
Mar 23, 2019

For most, number crunching can be a boring or tedious job. In such cases, having a trick or two up your sleeve will come in handy. Here, we'll determine how to find the square of a number ending in 1 or 5, with some useful pointers.

A square of any given number will always end in 0, 1, 4, 5, 6, or 9. This was evident in the table of squares that was published by mathematician Dr. Hutton, in the year 1781.

A math teacher always said that one must never allow math to be your master, but always be in a position that you are its master. But most of you would agree that this is easier said than done. To help us in becoming the masters of the tyrant named math, he had given us a few tips and tricks that sure do come in handy when dealing with numbers.

In geometry, two adjacent sides―one long and one short―of a rectangle are multiplied to find out the area of the rectangle. However, both the adjacent sides in a square are of the same size. Hence, the process of multiplying a number by itself is known as 'square'.

The following sections give you some great tricks to find the squares of numbers ending in 5, and also those numbers ending in 1.

Separate 5 in the units place from the number, and consider the number formed by the rest of the digits.

Multiply this number by its successive number.

Place the number 25 at the end of the product obtained.

The new number thus formed is the square of the original number.

Consider the number 45

**Step 1**

Take the digit in the tens place, which in this case is 4

**Step 2**

Multiply 4 by the number after 4, which is 5

Thus,

4 × 5 = 20

**Step 3**

Place the number 25 at the end of the product obtained, which is 20

We get 2025

The new number thus formed is 2025, which is the square of the given number 45

Take the digit in the tens place, which in this case is 4

Multiply 4 by the number after 4, which is 5

Thus,

4 × 5 = 20

Place the number 25 at the end of the product obtained, which is 20

We get 2025

The new number thus formed is 2025, which is the square of the given number 45

Find the square of 75

**Step 1**

The digit in the tens place is 7

**Step 2**

The number after 7 is 8

So, 7 × 8 = 56

**Step 3**

Place 25 at the end of the product 56, to get 5625

Thus, the square of 75 is 5625

The digit in the tens place is 7

The number after 7 is 8

So, 7 × 8 = 56

Place 25 at the end of the product 56, to get 5625

Thus, the square of 75 is 5625

Consider the number 145

**Step 1**

The number obtained from the digits in the tens and hundreds places is 14

**Step 2**

The number to succeed 14 is 15

Hence, 14 × 15 = 210

**Step 3**

Place 25 at the end of the obtained product 210, to get 21025

Thus, the square of 145 is 21025

The number obtained from the digits in the tens and hundreds places is 14

The number to succeed 14 is 15

Hence, 14 × 15 = 210

Place 25 at the end of the obtained product 210, to get 21025

Thus, the square of 145 is 21025

A two-digit number that ends in 5 can be represented as 10x + 5.

Square of this term is (10x + 5) × (10x + 5).

This can be represented as (10x + 5)^{2}.

Square of this term is (10x + 5) × (10x + 5).

This can be represented as (10x + 5)

Now, by using,

(a + b)^{2} = a^{2} + (2 × a × b) + b^{2}

We get,(10x + 5)^{2}= (10x)^{2} + (2 × 10x × 5) + 5^{2}

= 100x^{2} + 100x + 25

= 100x(x + 1) + 25

Here, the number obtained is x(x + 1)―25.

**Here, '―' is only a separator between the digits of a same number.*

(a + b)

We get,(10x + 5)

= 100x

= 100x(x + 1) + 25

Here, the number obtained is x(x + 1)―25.

Separate 1 in the units place from the number, and consider the number formed by the rest of the digits.

Obtain the square of this digit, and also its product with two.

Place the digits of the squared value followed by the digit in the units place of the product. If there is more than one digit in the product, then other than the digit in the units place, add the rest of them to the squared value.

Place the digit 1 at the end of the number obtained from the previous step. The number thus obtained is the square of the original number.

Consider the number 51

**Step 1**

Considering the number after ignoring 1: 5

**Step 2**

Square of 5 = 5^{2} = 25

5 × 2 = 10

Here, the product is a two-digit number. Hence, add all the digits other than the digit in the units place to the squared value.

Considering the number after ignoring 1: 5

Square of 5 = 5

5 × 2 = 10

Here, the product is a two-digit number. Hence, add all the digits other than the digit in the units place to the squared value.

The number formed by putting together results from Step 2 is (25 + 1)0 = 260

Placing one at the end of this number, we get 2601

2601 is the square of the given number 51

This is applicable to three-digit numbers as well.

Consider the number 151.

**Step 1**

Number formed by digits remaining after ignoring 1: 15

**Step 2**

Square of 15 = 15^{2} = 225

15 × 2 = 30

**Step 3**

The number formed by putting together the results from Step 2 is (225 + 3)0 = 2280

**Step 4**

Placing 1 at end of the number obtained in above step, we get 22801

Number formed by digits remaining after ignoring 1: 15

Square of 15 = 15

15 × 2 = 30

The number formed by putting together the results from Step 2 is (225 + 3)0 = 2280

Placing 1 at end of the number obtained in above step, we get 22801

A number that ends in one can be represented as (10x + 1).

Square of this term is (10x + 1) × (10x + 1).

(10x + 1) × (10x + 1) = (10x + 1)^{2}

Again, we use the identity

(a + b)^{2} = a^{2} + (2 × a × b) + b^{2}

to get,

(10x + 1)^{2} = (10x)^{2} + (2 × 10x × 1) + 1

= 100x^{2} +20x + 1

The number in this case is x^{2}―2x―1.

**Here, '―' is only a separator between the digits of a same number.*

Square of this term is (10x + 1) × (10x + 1).

(10x + 1) × (10x + 1) = (10x + 1)

Again, we use the identity

(a + b)

to get,

(10x + 1)

= 100x

The number in this case is x

The term 'square', used when a number is multiplied by itself, has been in use since the 16th century. These techniques given to find the square are not only easy, but also save a lot of time by decreasing the time spent in lengthy calculations.