**Application of Concept of Cyclicity**

__How to calculate unit digit if a number contains power of a power__

**Example 1:What will be the last digit of ${{12}^{{{23}^{45}}}}$**

**Solution:**

To find the last digit of this type of number we will start the question from the base the base is given to be 12. It means we will see the cyclicity of 2 because the last digit depends upon the unit digit of 12 i.e. 2. We observe unit digit while calculating powers of 2

2^{1} = 2

2^{2} = 4

2^{3} = 8

2^{4} = 6

2^{5} = 2

**Step 1**: Now we know that cyclicity of last digit of 12 i.e. 2 is of 4, hence we divide the power of 12 i.e. 23^{45} with 4.

**Step 2**: Now lets calculate the remainder of 23^{45 }when divided by 4 and then we will determine the last digit.

**Step 3**: The remainder will be 3 because we can write remainder of 23 by 4 as 3 or -1.Hence we can write 23^{45} as (-1) ^{45} but odd power of -1 will be again -1 and thus 23^{45 }when divided by 4 will give us remainder as -1 or 3.

Hence unit digit of **${{12}^{{{23}^{45}}}}$** will be same as unit digit of 2^{3}i.e. 8

**Example 2: ****Find the unit digit of ${{32}^{{{25}^{95}}}}$ _{ }**

**?**

**Solution: **To find the last digit of this type of number we will start the question from the base the base is given to be 32. It means we will see the cyclicity of 2 because the last digit depends upon the unit digit of 32 i.e. 2. We observe unit digit while calculating powers of 2

2^{1} = 2

2^{2} = 4

2^{3} = 8

2^{4} = 6

2^{5} = 2

**Step 1**: Now we know that cyclicity of last digit of 32 i.e. 2 is of 4, hence the divide the power of 12 i.e. 25^{95} with 4.

**Step 2**: Now lets calculate the remainder of 25^{95 }when divided by 4 and then we will determine the last digit.

**Step 3**: The remainder will be 1 because we can write remainder of 25 by 4 as 1. Hence we can write 25^{95} as (1) ^{95} but any power of 1 will be again 1 and thus 25^{95 }when divided by 4 will give us remainder as 1 .

Hence unit digit of **${{32}^{{{25}^{95}}}}$ _{ }** will be same as unit digit of 2

^{1 }i.e. 2

**Assignment:**

__Questions:__

**1:** Find the unit digit of 4^{686}

a) 4

b) 6

c) 2

d) 3

Answer: b

Solution: We know

4^{1} = 4

4^{2} = 6

4^{3} = 4

4^{4} = 4

Here, we see that powers of four repeat after a cycle of 2 i.e.

Odd power of 4 gives unit digit as 4 and even power of 4 gives unit digit as 6

So unit digit of 4^{686 }is 6.

**2 :** What will be the unit digit of 65^{1234}

a) 5

b) 0

b) 2

d) 3

Answer: a

Solution We know

5^{1} = 5

5^{2} = 25

5^{3} = 125

5^{4} = 625

So, we observe that unit digit of any power of 5 is 5.Hence unit digit of 65^{1234} is 5.

**3:** Find the unit digit of 36^{512}

a) 4

b) 6

c) 2

d) 3

Answer: b

Solution We know

6^{1} = 6

6^{2} = 36

6^{3} = 216

So, we observe that unit digit of any power of 6 is 6.Hence unit digit of 36^{512} is 6.

**4:** Find the unit digit of 18^{5693}

a) 4

b) 6

c) 2

d) 8

Answer: d

Solution: We observe unit digit while calculating powers of 8

Unit digit in 8^{1} = 8

Unit digit in 8^{2} = 4

Unit digit in 8^{3} = 2

Unit digit in 8^{4} = 6

Unit digit in 8^{5} = 8

So on dividing 5693 with 4, 1 will be the remainder and hence the last digit would be 8.