For the concept of identifying the unit digit, we have to first familiarize with the concept of cyclicity. Cyclicity of any number is about the last digit and how they appear in a certain defined manner. Let’s take an example to clear this thing:

The cyclicity chart of 2 is:

2^{1 }=2

2^{2} =4

2^{3 }=8

2^{4}=16

2^{5}=32

Have a close look at the above. You would see that as 2 is multiplied every-time with its own self, the last digit changes. On the 4^{th} multiplication, 2^{5} has the same unit digit as 2^{1}. This shows us the cyclicity of 2 is 4, that is after every fourth multiplication, the unit digit will be two.

Cyclicity table:

The cyclicity table for numbers is given as below:

How did we figure out the above? Multiply and see for yourself. It’s good practice.

Now let us use the concept of cyclicity to calculate the Unit digit of a number.

**What is the unit digit of the expression 4 ^{993}? **Now we have two methods to solve this but we choose the best way to solve it i.e. through cyclicity

We know the cyclicity of 4 is 2

Have a look:

4

^{1}=4

4

^{2 }=16

4

^{3 }=64

4

^{4 }=256

From above it is clear that the cyclicity of 4 is 2. Now with the cyclicity number i.e. with 2 divide the given power i.e. 993 by 2 what will be the remainder the remainder will be 1 so the answer when 4 raised to the power one is 4.So the unit digit in this case is 4.

For checking whether you have learned the topic, think of any number like this, calculate its unit digit and then check it with the help of a calculator.

* Note* :

*If the remainder becomes zero in any case then the unit digit will be the last digit of*

**a**^{cyclicity number }*where a is the given number and cyclicity number is shown in above figure.*

Lets solve another example:

**The digit in the unit place of the number 7 ^{295} X 3^{158} is**

A. 7

B. 2

C. 6

D. 4

**Solution**

The Cyclicity table for 7 is as follows:

7^{1 }=7

7^{2 }=49

7^{3} = 343

7^{4} = 2401

Let’s divide 295 by 4 and the remainder is 3.

Thus, the last digit of 7^{295} is equal to the last digit of 7^{3} i.e. 3.

The Cyclicity table for 3 is as follows:

3^{1} =3

3^{2} =9

3^{3 }= 27

3^{4 }= 81

3^{5 }= 243

Let’s divide 158 by 4, the remainder is 2. Hence the last digit will be 9.

Therefore, unit’s digit of (7^{925} X 3^{158}) is unit’s digit of product of digit at unit’s place of 7^{925} and 3^{158} = 3 * 9 = 27. Hence option 1 is the answer.

the value of 4^102 is supposed to have 6 at the unit place.Calculating by your method..4 has a cyclicity of 2. Dividing 102 by 2, we get remainder as zero. 4^0=1. Buh!

The answer is 6

In case of zero remainder the unit digit will be the last digit of a^cyclicity number

Here the value of a is 4 and the value of cyclicity number is 2

Hence the last digit of 4^102 will be 6

what willbe the answer of 2^4 with ur logic i will get 2 thats wrong answer will be 6

The answer will be 6

When we divide 4 with the cyclicity number i.e 4 ,the remainder will be zero

and in case of remainder zero the last digit of the number will be last digit of 2^cyclicity number

Now the cyclicity number of 2 is 4

Therefore the unit digit of 2^4 is 6

this topic is very useful for all students

for example 0 is the remainder but 4 power of 0 is not there .my question is 4^44 .we got the remainder 0 so we 4^0=1 but 1 is not a correct answer .

please explain the answer.

Sir in case of zero remainder,the unit digit will be the last digit of 4^cyclicity number

As the cyclicity of 4 is 2 therefore the in case of zero remainder the unit digit will be 6 not 4^0 =1

for example

4^2 = last digit is 6

4^4 = last digit is 6

4^6 = last digit is 6