Unit digit is the last digit of any number from right to left. As for example in 123 right most digit is 3, so unit digit is 3. Similarly in 5^{3} = 125 unit digit is 5. But all the time you cannot expand x^{y} to a flat number as you easily do in case of 5^{3} = 125. In this article we are going to discuss how will you find unit digit of x^{n} like 2017^{2018} or 144^{145} × 126^{126} within seconds.

To solve problems like this we need to invest only some minutes to understand the funda behind it. We just need to have the idea of cyclicity of a number. Cyclicity means when the unit digit( repeats. Didn’t get my point? Don’t worry. Just follow me…

## Cyclicity of 0

In 0^{n} unit digit will be always 0

## Cyclicity of 1

In 1^{n} unit digit will be always 1

## Cyclicity of 2

2^{1}=2

2^{2}=4

2^{3}=8

2^{4}=16

2^{5}=32

2^{6}=64

2^{7}=128

2^{8}=256

2^{9}=…2

2^{10}=…4

………

Look in 2^{9} and 2^{10} we didn’t write the total expansion as we are only concerned with unit digit. Notice that 2, 4, 8, 6 repeats as a unit digit. Total repeating numbers are 4. So we say that cyclicity of 2 is 4. Now to find the unit digit of 2^{n} we simply divide n by 4 and the remainder will tell about the unit digit.

Q. Find the unit digit number of 2^{54}

So unit digit of 2^{54} is 4

Q. Find the unit digit number of 2^{223}

So the unit digit number of 2^{223} is 8

## Cyclicity of 3

3^{1}=3

3^{2}=9

3^{3}=27

3^{4}=81

3^{5}=243

3^{6}=…9

3^{7}=…7

3^{8}=…1

3^{9}=…3

Here 3, 9, 7, 1 repeat. So cyclicity is 4

Q. Find the unit digit number of 3^{223}

So unit digit of 3^{223} is 7

Q. Find the unit digit number of 223^{223}

Here I want to mention you one thing that you need to focus only on unit digit of x. If we write the problem in x^{n} form x=223, unit digit=3

So unit digit number of 223^{223} is 7.

**Note: We can conclude from the above two problems that for power n to various x that have the same unit digit will generate an end number with same unit digit.**

## Cyclicity of 4

4^{1}=4

4^{2}=16

4^{3}=64

4^{4}=256

Here 4 and 6 are repeating numbers. So Cyclicity is 2. To make it simple we can remember that is unit digit of 4^{n} is 4 when n is odd number and 6 when n is even number.

Q. Find the unit digit number of 204^{223}

n=223→it’s an odd number

So unit digit is 4

Q. Find the unit digit number of 963^{369} × 204^{223}

Just find the unit digit of both the numbers and multiply. Product’s unit digit is required unit digit

963^{369}\times 204^{223}= 3^{369}\times 4^{223}=3\times 4 = 12 =2So unit digit of 963^{369} × 204^{223} is 2

## Cyclicity of 5

5^{1}=5

5^{2}=25

5^{3}=125

5^{4}=625

Unit digit will be always 5

## Cyclicity of 6

6^{1}=6

6^{2}=36

6^{3}=216

6^{4}=…6

So unit digit will be always 6

## Cyclicity of 7

7^{1}=7

7^{2}=49

7^{3}=343

7^{4}=…1

7^{5}=…7

7^{6}=…9

7^{7}=…3

7^{8}=…1

7^{9}=…7

7, 9, 3, 1 repeat. So Cyclicity is 4

Q. Find the unit digit number of 963^{369} × 204^{223} × 2017^{201}

So unit digit of 963^{369} × 204^{223} × 2017^{201 }is 4

## Cyclicity of 8

8^{1}=8

8^{2}=64

8^{3}=512

8^{4}=…6

8^{5}=…8

8^{6}=…4

8^{7}=…2

8^{8}=…6

8^{9}=…8

Here 8, 4, 2, 6 repeat. So Cyclicity is 4

Q. Find the unit digit number of 18^{18} × 28^{28} × 2018^{2019}

18^{18}\times 28^{28}\times 2018^{2019}

= 8^{18}\times 8^{28}\times 8^{2019}

= 8^{4k_1+2}\times 8^{4k_2+0}\times 8^{4k_3+3} [when 1^{st} and 3^{rd} powers are divided by 4, remainders are 2, 3 respectively but 2^{nd} power is multiple of 4, so we end up in 8^{4 }while applying cyclicity]

= 8^{2}\times 8^{4}\times 8^{3}

= 4\times 6\times 2

= 48

= 8

∴ Unit digit is 8

**Question**. Find the unit digit number of 18^{18}\times 28^{28} \times 2018^{2019}

(a) 8

(b) 4

(c) 2

(d) 6

**Ans**: (a) 8

**Explanation**:

18^{18}\times 28^{28}\times 2018^{2019}

= 8^{18}\times 8^{28}\times 8^{2019}

Now, 8^{18}\times 8^{28}\times 8^{2019}

= 8^{4k_1+2}\times 8^{4k_2+0}\times 8^{4k_3+3} [when 1^{st} and 3^{rd} powers are divided by 4, remainders are 2, 3 respectively but 2^{nd} power is multiple of 4, so we end up in 8^{4 }while applying cyclicity]

= 8^{2}\times 8^{4}\times 8^{3}

= 4\times 6\times 2

= 48

= 8

So, we have got unit digit = 8

## Cyclicity of 9

9^{1} = 9

9^{2} = 81

9^{3} = 729

9^{4} = …1

9^{5} = …9

Hence, when power is odd, unit digit = 9 and when power is even, unit digit = 1

**Final Note**: You actually don’t require to memorize cyclicity of any number. By practice it will definitely stick to your mind. Anytime you feel doubt just check it, it’s only limited to 4 power max. By following the link provided below practice some problem, so that getting unit digit of a number becomes a piece of cake to you.

## Concept of negative remainder to find unit digit

Actually unit digit is the remainder when any number is divided by 10. So, to make a negative remainder positive 10 need to be added with it. See the following example when 29 is divided by 10 to get the unit digit 9.

**Question**: What is the digit in the unit place of 24^{63}+33^{61}-27^{58}

**Question**: If x=164^{169}+333^{337}-727^{726}, then what is the unit digit of x?

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