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

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 below. 

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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