Concept of Last Two Digits

Hello All,
First of all let me appreciate the opportunity of being a part of this team. This is my first article to this project.
Before I start the concept of last two digits lets have a quick look at the concept behind unit’s digit. I know most of you might be aware of this concept called cyclicity. All you need to do, is just catch the last digit of the base and apply the rule of cyclicity. You can follow the cyclicity table give below for your convenience.
Base Index-1 Index-2 Index-3 Index-4
Type 4K+1 4K+2 4K+3 4K
1 1 1 1 1
2 2 4 8 6
3 3 9 7 1
4 4 6 4 6
5 5 5 5 5
6 6 6 6 6
7 7 9 3 1
8 8 4 2 6
9 9 1 9 1
0 0 0 0 0

e.g Last digit or unit digit of 127^127.(Please note that this expression is to be read as 127 to the power 127. Also all other similar expressions are also to be followed in the same pattern.)
Just pick up last digit of the base i.e 7 and divide the given power by 4 & check out the remainder.
The remainder when 127 is divided by 4 is 3. Hence the power of the given expansion is of the type 4K+3. Therefore the unit’s digit will be 3.
How simple is that.
Now lets move on to the concept of last two digits. The fundamental behind the last two gits is just like unit’s digit concept.
Here you just need to catch the last two digits.
e.g Find the last two digits of the product of 1023 X 2027 X 5096 X 5979.
Just pick up the last two digits of all the numbers.
i.e 23, 27, 96 & 79
Multiply any pair using crisscross method of multiplication and note down the last two digits only.
i.e 23 X 27 gives you 21 as last two digits similarly 96 x 79 gives you 84 as last two digits.
Now again multiply the obtained results to get the answer.
i.e 21 x 84 which gives you 64 as last two digits.
Hope you find it doable comfortably.
Now lets move on to the cyclicity concept of last two digits.
1. If you have 1 as your unit’s digit in any base then whatever be the ten’s place, you just need to write 1 as your unit’s place as simple as that & just multiply the unit digit of index/power to the ten’s place of the base and note down the unit digit of multiplication as your ten’s digit of expansion.
e.g Last two digits of 21^144.
You have 1 as unit’s place and 2 as ten’s place in your base in the given example.
You know that unit digit of this expansion has to be 1 always.
Now for ten’s digit of this expansion you pick up the unit digit of the power 144, which is 4 & multiply it to the ten’s digit of the base which is 2.
i.e 4 X 2 which will be 8 & this will be the ten’s place of the expansion.
Hence last two digits come out to be 81.
Similarly 3, 7 & 9 can be converted to such an extent so that you can get unit digit 1 and after that you can easily do the process of 1 as unit’s place.
One more very interesting and very important result is the cyclicity of 24 as last two digits of the base.
Please follow if 24^even power it will always end by 76 as last two digits
& if 24^odd power, it will always end by 24 as last two digits.
Also 2, 4 & 6 can be broken into factors to get the last two digits.
e.g Last two digits of 196^224.
The base 196 can be converted to small factors like 196 = 4 X 49
Now the problem becomes simple 4^224 x 49^224

Or (4^5)^44 X 4^4 X 2401^112

which gives you 76, 56 & 01 as last two digits respectively.
Multiply step by step as the process discussed earlier and get the last two digits.
The answer will be 56 for this example.
Concept of compliment of the base
e.g Last two digits of 39^44 .
Compliment of the base will be 61, hence lets take out last two digits of (-61)^44 which results to -41 as last two digits. But this has to be positive and to make it positive just add the complimentary base i.e 100 to the last result -41. Hence the answer comes out to be 59.
And lastly when 5 is your unit’s place in the base. It can be of two forms.
Either ten’s place is odd or ten’s place is even.
i.e either of the type 15, 35, 55, 75, 95, (Lets say Case-1)
or of the type 25, 45, 65, 85. (Lets say Case-2)
Case-1 last two digits will be 75 for any odd power .
And for any even power last two digits will be 25.
Case-2 Last two digits will be 25 for any power .
Now you can have any example of the types discussed in this article and easily take out the last two digits of the same.
Thats all for now. I believe that you all will be able to understand the concept of last two digits from this article. I will be waiting for your responses and valuable suggestions.
All the best
Rajneesh Tripathi

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I welcome the opportunity to help the students who are willing to write CAT,XAT & many more similar competitive entrance exams. I have been teaching to similar category of students since last six years. during my transition I have worked with IMS & TIME and trained a number of students over the period. I wish all the viewers a very happy learning experience through my contents posted.

Rajneesh Tripathi

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