HOW TO FIND UNIT DIGIT OF A NUMBER WITH POWER

To identify the unit digit of a number with some power, we must be aware of cyclicity.

Cyclicity of Numbers

Cyclicity of any number is about the last digit and how they appear in a certain defined manner. 

Example 1 :

Let us consider the values of 2ⁿ, where n = 1, 2, 3, ...........

2¹  =  2

2²  =  4

2³  =  8

2⁴  =  16

2⁵  =  32

2⁶  =  64

In the above calculations of 2ⁿ, 

We get unit digit 2 in the result of 2ⁿ, when  n  =  1.

Again we get 2 in the unit digit of 2ⁿ, when n  =  5.

That is, in the fifth term. 

So, the cyclicity of 2 is 4.

Example 2 :

Let us consider the values of 3ⁿ, where n = 1, 2, 3, ...........

3¹  =  3

3²  =  9

3³  =  27

3⁴  =  81

3⁵  =  243

3⁶  =  729

In the above calculations of 3ⁿ, 

We get unit digit 3 in the result of 3ⁿ, when  n  =  1.

Again we get 3 in the unit digit of 3ⁿ, when n  =  5.

That is, in the fifth term. 

So, the cyclicity of 3 is 4.

In the same way, we can get cyclicity of others numbers as shown in the table below. 

Example 1 :

Find the unit digit of 32²⁴.

Step 1 : 

Take the unit digit in 32 and find its cyclicity.

The unit digit of 32 is '2' and its cyclicity is 4.

Step 2 : 

Divide the exponent 24 by the cyclicity 4.

Step 3 :

When 24 is divided by 4, the remainder is zero. 

Because the remainder is zero, we can get the unit digit of 2⁸, from the last value of the cyclicity of 2ⁿ_.

The last value of the cyclicity of 2ⁿ is 

2⁴  =  16

The unit digit of 16 is '6'. 

Therefore, the unit digit of 32²⁴ is 6.

In step 3 of the above example, what if the remainder is not zero?

It has been explained in the next example. 

Example 2 :

Find the unit digit of 32²⁷.

Step 1 : 

Take the unit digit in 32 and find its cyclicity.

The unit digit of 32 is '2' and its cyclicity is 4.

Step 2 : 

Divide the exponent 27 by the cyclicity 4.

Step 3 :

When 27 is divided by 4, the remainder is 3. 

Take the remainder 3 as power of 2 (unit digit of 32). 

2³  =  8

Therefore, the unit digit of 32²⁷ is 8.

Solved Questions

Question 1 :

Find the  unit digit of (3547)¹⁵³_.

Solution :

In (3547)¹⁵³, unit digit is 7.

The cyclicity of 7 is 4. Dividing 153 by 4, we get 1 as remainder.

7¹  =  7

Therefore, the unit digit of (3547)¹⁵³ is 7.

Question 2 :

Find the unit digit of (264)¹⁰²_.

Solution :

In (264)¹⁰², unit digit is 4.

The cyclicity of 4 is 2. Dividing 102 by 2, we get 0 as remainder.

Since the remainder is 0, unit digit will be the last digit of a cyclicity number.

4²  =  16 (the unit digit is 6)

Unit digit of 4¹⁰² is 6.

Therefore, the unit digit of (264)¹⁰² is 6.

Question 3 :

What is the unit digit in the product

(7)¹⁰⁵

Solution :

The unit digit is 7.

The cyclicity of 7 is 4. Dividing 105 by 4, we get 1 as remainder.

So, 

7¹  =  7

The unit digit is 7.

Therefore the unit digit of (7)¹⁰⁵ is 7.

Question 4 :

Find unit digit of (3⁶⁵ x 6⁵⁹ x 7⁷¹). 

Solution :

Cyclicity of 3 is 4. Dividing 65 by 4, we get the remainder 1.

Then,

3¹  =  3

So, the unit digit of 3⁶⁵ is 3.

Cyclicity of 6 is 1. Dividing 59 by 1, we get the remainder 0.

Then,

6¹  =  6

So, the unit digit of 6⁵⁹ is 6.

Cyclicity of 7 is 4. Dividing 71 by 4, we get the remainder 3..

Then,

7³  =  343

The unit digit of 343 is '3'. 

So, the unit digit of 7⁷¹ is also 3.

Product of unit digits : 

  =  3 x 6 x 3

  =  54

The unit digit of the product is 4.

Therefore, the unit digit of (3⁶⁵ x 6⁵⁹ x 7⁷¹) is 4.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.