# How to find out the last digit of a number raised to a high power? Hello CAT aspirants

Many students find this topic confusing. Well, time to chuck the confusion out of the window. Let’s begin! Cyclicity is defined as the quality or state of being cyclic. Cyclicity of a number is the number of times after which the number repeats itself in a pattern. In this article, we will focus on cyclicity related to last digit pattern in numbers raised to high power. For instance, let us focus on the increasing powers of 2 and just check out their last digit.

2^1 = 2 = 2^5 = 2^9

2^2 = 4 = 2^6

2^3 = 8 = 2^7

2^4 = 6 = 2^8

If you look at the pattern, Last digit of 2^1, 2^5, 2^9 and so on are the same. Look at the powers after which repetition happens: 5-1=4. 9-5=4. This implies cyclicity for last digit in case of 2 is 4. Which means, Last digit of 2^(4n+1) will be 2. 2^(4n+2) will end with 4, 2^(4n+3) will end with 8 and 2^(4k+0) will end with 6. So, all we need to check is “What is the remainder when power is divided by 4?”

Let’s take some examples:

2^42 => Divide 42 by 4 => 42 = 4*10 + 2 => This is of the form 4n+2 => Power will be 2. => 2^2 = 4.

2^53 => Divide 53 by 4 => 53 = 4*13 + 1 => This is of the form 4n+1 => Power will be 1. => 2^1 = 2.

2^48 => Divide 48 by 4 => 48 = 4 *12 + 0 => This is of the form 2^4k => 2^4 = Last digit = 6.

Similarly check for cyclic patterns in the last digit for other digits. This is what you will find:

Digits when Cyclicity is 4: One last example on this: Find the last digit of 28^34

Rule: When finding out the last digit, work only with the last digit. So, the question is: 8^34

Now, cyclicity of 8 is 4. So, Divide power by 4. => 34 = 4*8+2 => Power is 4n+2 => 8^2 = 4. Hope it is clear.

Let’s now take a case when there is power on power:

Find the last digit of 12^13^14

Now, rule remains the same: Divide 13^14 by 4 to find the remainder. (13)^14 % 4 = (4*3 + 1)^14 % 4 = 1^14 = 1 => Remainder = 1. => 2^1 = 2.

Now, coming to digits when Cyclicity = 2: Digits 4 and 9

Let’s take a case first: Let’s take last digit as 9 and check only the last digit.

9^1 = 9 = 9^3 =9^5

9^2 = 1 = 9^4

See now! Repetition is happening at a difference of 2. Hence in such cases all we need to check is whether the power is odd or even. Here comes the table: Let’s take some examples: Find the last digit of 4^56

Just check the power: 56 is even => 4^even = 6.

Next example: Find the last digit of 9^8^7

This case has power on power => Look at the power => 8^7 => even^7 = even.

Hence 9^even = 1. Hope this is clear!

Final case: When cyclicity = 1

Digits: 0,1,5,6 => These digits repeat every time. Let’s take some examples:

5^1 = 5, 5^2 = 5, 5^3 = 5……last digit is always 5. Same with 0,1,6.

Ex: 21^12 = 1^12 = 1

Ex: 15^6 = 5^6 = 5.

So, in numbers ending with 0,1,5,6 always end with the same digit 0,1,5,6 irrespective of the power given.

So,finally let’s come to the question we started with:

Solution: Digit is 2 => Cyclicity is 4 => Power will be divided by 4.

Power = 2^2^2 = 2^4 = 16 => Remainder when 16 is divided by 4 = 0 => Power is of the form 4k.

=> 2^4k => 6

Simple! Happy learning chaps!

Will share more articles on Quants in due course!

Rahul Sir

Teacher, Bfactory