HOW TO FIND THE UNIT DIGIT IN THE PRODUCT

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 below. 

Cyclicity of 4 and 5

Cyclicity of 4

41  =  4

42  =  16

43  =  64

44  =  256

The cyclicity of 4 is 2.

Cyclicity of 5

51  =  5

52  =  25

53  =  125

The cyclicity of 5 is 1.

Cyclicity of 6 and 7

Cyclicity of 6

61  =  6

62  =  36

63  =  216

The cyclicity of 6 is 1.

Cyclicity of 7

71  =  7

72  =  49

73  =  343

74  =  2401

75  =  16807

The cyclicity of 7 is 4.

Cyclicity of 8, 9 and 10

Cyclicity of 8

81  =  8

82  =  64

83  =  512

84  =  4096

85  =  32768

Cyclicity of 8 is 4.

Cyclicity of 9

91  =  9

92  =  81

93  =  729

Cyclicity of 9 is 2.

Cyclicity of 10

101  =  10

102  =  100

Cyclicity of 10 is 1.

Cyclicity of Numbers - Summary

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

Solved Problems

Example 1 :

Find the unit digit in the product :

(3547)153 x (251)72

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

So, the unit digit of 7153 is 7.

In 25172, unit digit is 1.

Because 1 has the cyclicity 1, the unit digit of 25172 is 1.

By multiplying the unit digits, we get

7 x 1  =  7

Therefore, the unit digit of the expression

(3547)153 x (251)72 is 7.

Example 2 :

Find unit digit in the product :

(6374)1793 x (625)317 x (341)491

Solution :

In (6374)1793, unit digit is 4.

The cyclicity of 4 is 2. Dividing 1793 by 4, we get 1 as remainder.

41  =  4

So, the unit digit of (6374)1793 is 4.

In (625)317, unit digit is 5.

Since 5 has the cyclicity 1, the unit digit of (625)317 is 5.

In (341)491, unit digit is 1.

Since 1 has the cyclicity 1, the unit digit of (341)491 is 1.

By multiplying the unit digits, we get

4 x 5 x 1  =  20

The unit digit of 20 is '0'.

Therefore, the unit digit of the expression

(6374)1793 x (625)317 x (341)491 is 0

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