# CAT Prep Series : How to find out last two digits of a number

I hope that the title of the post answers the question, “what is this particular post about?” very clearly. What it might not answer is – “why is it even relevant?” Before I go ahead to answer both those questions, let me tell you a little bit about this series. I am Ravi Handa, founder of **Handa ka funda**, and I have been running an online CAT coaching course for past three years. Prior to that, I worked at IMS for five years. **MISB Bocconi** is the world’s only offshore presence of Università Bocconi, one of Europe’s premier universities. When the MISB team contacted me about a few blog posts, we tried to figure out the best possible way forward in which we can help a CAT aspirant. And this is the result of what we have decided. We plan to have a series of 6 articles before 29^{th} November, the day on which CAT 2015 is scheduled, in which we will try to cover a diverse range of topics such as ‘How to Improve Reading Comprehension Skills’ to ‘What to do in one week before the CAT 2015 exam?’ Your feedback and criticism are requested so that we can make it better and make this series more helpful to a CAT aspirant.

With the thought process of this exercise out of the way, let me start with the topic at hand.

Finding out last two digits of a number is a skill that will help you not only solve some tough questions but quickly eliminate a few wrong options as well. Also, it is a skill that you can develop and master in less than 15 minutes. All you need to do is to remember a few rules and the rest will follow.

**Funda 1: A number ending in 1**

*Last two digits of (…a1)^(…b) will be [Last digit of a*b]1*

Ex 1.1 Last two digits of 491^83 = [Last digit of 9*3]1 = [Last digit of 27]1 = 71

Ex 1.2 Last two digits of 571^64 = [Last digit of 7*4]1 = [Last digit of 28]1 = 81

** ****Funda 2: Solving last two digits for odd numbers**

*Change the odd number to something that ends in 1.*

Ex 2.1 What are the last two digits of (86789)^41?

For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.

After that, we can use the property, last two digits of (…a1)^(…b) will be [Last digit of a*b]1

Let us try and apply this concept in the given question

Last two digits of (86789)^41

= Last two digits of 89^41

= Last two digits of 89 * 89^40

= Last two digits of 89 * (..21)^20

= Last two digits of 89 * 01 {Here I have used the concept mentioned above} = 89

**Funda 3: Solving last two digits for even numbers**

*24 raised to an odd power will end in 24 and even power will end in 76*

This comes in handy when you are calculating last two digits of an even number or a power of 2. Do keep in mind that 2^10 is 1024

Let me take an example to explain this concept further

Ex 3.1 Find last two digits of 1456^72

1456^72 = 16^72 x 91^72

Last two digits of 16^72 = Last two digits of 2^208 = Last two digits of 2^20 x 2^8 = Last two digits of 1024^2 x 256 = Last two digits of 76 x 56 = 56

Last two digits of 91^72 = [Last digit of 9*2]1 = [Last digit of 18]1 = 81

Overall the last two digits = Last two digits of 56 x 81 = 36

The point to note here is that I converted the given even number into powers of two and an odd number and then solved it.

**Funda 4: Last two digits of a number ending in 5**

*If the second last digit of the base and the power, both are odd – it will end in 75; otherwise the last two digits will be 25.** *

Ex 4.1 Last two digits of 75^65 = 75 (Second last digit of the base and the power – both odd)

Ex 4.2 Last two digits of 65^75 = 25

Ex 4.3 Last two digits of 35^73 = 75 (Second last digit of the base and the power – both odd)

Ex 4.4 Last two digits of 1995^2014 = 25

Ex 4.5 Last two digits of 1995^2015 = 75 (Second last digit of the base and the power – both odd)

I hope you would have learned something from this post about Number Theory, which forms an integral part of CAT Preparation. Do provide feedback about the same via the comments section on the blog. I look forward to your suggestions.

*Ravi Handa, an alumnus of IIT Kharagpur, has been teaching for CAT and various other competitive exams for around a decade. He started online courses on his website** **Handa Ka Funda** **in 2013 and 10000+ students have subscribed to them since then.*

Did not unf=derstand for the odd no.s :

“= Last two digits of 89 * (..21)^40”

Permalinkwhat did you do.

Nice article… will certainly be helpful.. please share the method of finding the last non-zero digit of any number..

PermalinkEx 4.1 Last two digits of 75^65 = 75 (Second last digit of the base and the power – both odd)

Here second last digit of power ie 65 is not ODD. It is even. Are you talking about only last digit in case of power?

Permalink@Anmol – The concept that I have used is, if you have an odd number ending in 1 – you can predict its last two digits very easily by just multiplying the second last digit of the base with the last digit of the power.

For example

Last two digits of 89 * (..21)^40

Will be the last two digits of 89 * (..01)

How did I get (..01)?

Well, second last digit of the base is 2

Last digit of the power is 0

Their product is 2*0 = 0

So, it leads to 01

Let us look at some more examples

Last two digits of 76421^23133 = 61 (because 2*3 = 6)

Last two digits of 71^48 = 61 (because 7*8 = 56 – we will just need 6)

Hope this clarifies.

Permalink@Soumitra – Thanks for the feedback. I am not sure if I understand your question. Can you please give me an example of what you are looking for. I will surely look into it in one of the future posts.

Permalink@Tuhar – Yes. I am talking about the power (not the second last digit of the power)

PermalinkHello Sir, can you please explain how 89*(89^40)=89*(..21)^40. Thank you so much for your help.

Permalinkit’s a typo…he wrote 40 instead of 20

89*(89^40) = 89*(7921^20) = 89*01 = 89

Permalink16^72=2^288….plz clarify that solution

Permalink@Tushar – You are right. That is indeed a typo. But that will not change the final answer.

89*(89^40)= 89*(..21)^20 = 89*01 = 89

@Noel – 16^72=2^288 = (2^4)^72 = 2^(4*72) = 2^288

PermalinkSir, how come “Last two digits of 2^208 = Last two digits of 2^20 x 2^8”

PermalinkPlease explain