Finding the last two digits of a^x form numbers .

You must have encountered questions asking to find the remainder when 453²⁷ (just an example) is divided by 100.
What has been your approach till now?
Was it by breaking 100 as 25*4 and then applying Chinese remainder theorem?
If yes, then go through this good alternative for such questions.

Finding the remainder when divided by 100 is nothing but finding the last two digits.
Like when 5³ is divided by 100, the remainder is 25 and 5³ = 125 and 25 is the last two digit.
So basically you should approach it like this way, and not by CRT.

Let’s start then:
We will deal with it in three parts
1. Numbers ending in 2/4/8 (2^k form)
2. Numbers ending in 1/3/7/9
3. Numbers ending in 6/5

Numbers ending in 2/4/8

Two things to remember before we start
1) The last two digits of 24^even will always be 76.
2) The last two digits of 24^odd will always be 24.

Let’s begin:
Whenever you see number ending in 2/4/8, try converting them into (..24)^x form. How will we get that?
Using 2¹⁰, as it is 1024.

Say, what are the last two digits of 2⁴⁵?
Try to convert it in (2¹⁰)^k form first and then separate the extra part.

45 = 40 + 5 = 10 × 4 + 5

So
2⁴⁵
= 2⁴⁰ × 2⁵
= (2¹⁰)⁴ × 2⁵ {Converted in (2¹⁰)^k and separated extra part is 2⁵}
= (…24)⁴ × 32
= 76 × 32 {∵ 24^even = …76}
= …32, the answer.

Let’s take another example: 4⁸⁷
First convert in 2^k form and then the same process.
4⁸⁷ = (2²)⁸⁷ = 2¹⁷⁴

Same process as above now
(Leaving for you to try from here)

Couple of Qs to try …
8⁴⁷⁵ and 16³³
Write the answers as comments.

Numbers ending in 1/3/7/9

Remember: Always try to convert the base number to end in a 1
How will it be done?
Since the cyclicity of numbers 1/3/5/7 is 4, and for each of these numbers, base⁴ will have 1 as unit digit, so try converting them to power of 4.
And after this say the last two digits are (A1)ⁿ …
… the unit digit is 1, and
…  the ten’s digit will be the last digit of A × n.

E.g. 23⁴¹
= (23⁴)¹⁰ × 23 (converted (a3)⁴ form)
= 41¹⁰ × 23
= 01 × 23 {4 × 0 = 0, so tens digit is 0}
= 23

One more to make it crystal clear

37⁸⁹
= 37⁸⁸ × 37
= (37⁴)²² × 37
= 61²² × 37
= 21 × 37 {6 × 2 = 12, so 2 is the ten’s digit}
= 77, the answer.

Numbers ending in 6/5

For number ending in 6 you have to basically separate it and solve.
I mean, say, 36⁴⁷
Using 36 = 9 × 4,
36⁴⁷ = 9⁴⁷ × 4⁴⁷

Now, proceed as above and solve.

For numbers ending in 5 just remember the following …
Say, the form is (x5)^y …
… whenever both x and y are odd, then the last two digits will be 75
… for all other cases it will be 25.

Couple of Qs to make it clear.

15³⁵
Both 1 and 35 are odd. Hence the last two digits is 75.

45³⁵
Both 4 and 35 aren’t odd. Hence the last two digits is 25.

Hope everything is clear now 🙂

Bonus tips:
1. Suppose in exam when solving 2ⁿ type Qs you are not able to remember whether 24^odd is 24 or 76, think 24¹ = 24, the power 1 is odd, so 24^odd is 24.
2. When not able able to recall last two digits of ((odd)5)^even. Think 15² = 225.

So basically all I want to say is taking small examples might help.

Hope it’s super sorted now 🙂