Finding Last Two digits

Concepts Examples

There can be direct questions on finding last two digits of any power of a natural number and this is very important for eliminating options in questions. Following are the four simple cases with which we will be able to solve almost every problem.

Rules for Finding Last Two Digits of square of a number

In order to find the last two digits of square of any two digit number number, we can write it as difference or sum from 50 or 100, whichever is closer. In general the last two digits of a number (50±k)² or (100-k)² will be determined by square of k.

e.g 47² can be written as (50-3)²,

Now, the last two digits will be determined by square of 3 that is 09

Rules for Finding Last Two Digits of any power of a number

1. For a natural number N^k, where N is a natural number ending with 0 and K is a natural number greater than 2,
   
   Last two digits will always be 00.

2. For a natural number N^k, where N is a natural number ending with 5 and K is a natural number greater than 2,
   
   Last two digits will always be 25.

3. For a natural number (2×m)^40k+1, where m is a odd number not ending with 5 and K is a natural number,
   
   Last two digits will always be (2×m +50).

4. For all the remaining cases, the last two digits of N^40k+x,
   
   Last two digits will always be equal to the last 2 digits of N^x

Question 1 Find the last two digits of 12345⁹⁸⁷⁴⁵.

05	00
25	50

C

Since the number ends with 5, thus rule 2 applies here

∴ the last two digits will be 25.

Question 2 Find the last two digits of 39².

21	81
79	67

A

Since 39² can be written as (50-11)²

∴ last two digits will be determined by 11²=21

Question 3 Find the last two digits of 22⁴⁸².

84	48
92	64

A

Since no rule applies to this, thus we will apply rule 4 here

22⁴⁸² can be written as 22^40×12+2

∴ the last two digits can be determined by 22² and will be 84

Question 4 Find the last two digits of 42⁸⁰¹.

29	92
45	54

B

Since the number can be written as (2×21)^40×20+1, thus rule 3 applies

∴ the last two digits will be (2×21+50)=92

Question 5 Find the last two digits of 39⁹⁰⁰.

01	81
27	34

A

Since no rule applies to this, thus we will apply rule 4 here

39⁹⁰⁰ can be written as 39^40×22+20

∴ the last two digits can be determined by 39²⁰



Now we will keep squaring 39 to reach closet to 39²⁰, we will consider only the last two digits.

39²- Last two digits will be 21, (using rule to find last two digits of square of a number (50-11)²)

39⁴- Last two digits will be 41, squaring 21

39⁸- Last two digits will be 81, squaring 41 (using rule to find last two digits of square of a number (50-9)²)

39¹⁶- Last two digits will be 61, squaring 81 (using rule to find last two digits of square of a number (100-19)²)



Now, last two digits of 39²⁰=Last two digits of (39⁴×39¹⁶)

=Last two digits of (41×61)

=01

∴ last two digits of 39⁹⁰⁰=01

Question 6 Find the last two digits of 53².

01	02
03	04

D

Since 81² can be written as (50+2)²

∴ last two digits will be determined by 2²=04

Question 7 Find the last two digits of 81².

21	61
27	67

B

Since 81² can be written as (100-19)²

∴ last two digits will be determined by 19²=61

Question 8 Find the last two digits of 80⁷⁵⁴.

64	16
00	08

C

Since the number ends with 0, thus rule 1 applies here

∴ the last two digits will be 00.