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, ...........

2 =  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. 

Number

1

2

3

4

5

6

7

8

9

10

Cyclicity of a number

1

4

4

2

1

1

4

4

2

1

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.

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