In general, we can find out that using binomial theorem.
The binomial theorem states that
Where
Let us see an example.
Suppose, we need to find the last two digits of 41876.
Using binomial expansion, we can write 41876 as
The first term in this expansion is 876C0 × 1876 =1
The second term of expansion = 876C1 × 1875 × 40 = 876 × 1 × 40 = 35040
The units digits of this term is 0 and tens digit will be the unit digit of product of 876 and 4, that is 4.
The third digit of the expansion =876C2 × 1874 × 402
Since 402 is there, the unit digit and tens digit of this term will be 0.
Similarly, the next all terms will be having unit digit as 0 since powers of 40 is involved.
So, the last two digits of 41876 is decided by the last two digits of the sum of first and second term.
Sum of first and second term =1+35040=35041
So, the last two digits of 41876 will be 41.
We can see that this a quite tedious process and so a short cut method is to be learned to solve this.
Short Cut Method to Find the Last Two Digits of a Number
The shortcut method is different for different numbers.
Last two digits of numbers ending in 1
For a number of the form xy where x ends in 1 like 2134, 1145, 3176, 10156, 102157 etc., the two steps given below are followed to find the last two digits.
For example: 21355
Tens digit =2×5=10
So, the tens digit will be 0.
And the unit digit is 1.
So, the last two digits will be 01.
Last two digits of numbers ending in 3,7 or 9
For a number of the form xy where x ends in 3,7 or 9 like 21334, 1745, 3976,?103?^56,?1029?^157 etc., the three steps given below are followed to find the last two digits.
For example: 19266
19266 = (192 )133
Now, 192=361 and ends in 61
Therefore, we need to find the last two digits of (361)133.
Once the number ends in 1, we can straight away get the last two digits with the help of the previous method.
Now, the tens digit is the product of tens digit of x and unit digit of y.
So, tens digit =8 (6 × 3 = 18)
The unit digit is 1.
Hence, the last two digits are 81.
Last two digits of numbers ending in 2,4,6 or 8
There is only one even 2-digit number which always ends in itself-76. That is, 76 raised to any power gives the last two digits as 76. Therefore, we should make 76 as the last two digits for even numbers.
We know that 242=576 and ends in 76 and 243=13824 and ends in 24.
24 raised to an even power always ends with 76, and 24 raised to an odd power always ends with 24
For example: 2434 will end in 76 and 2453 will end in 24.
Now those numbers which are not in the form of 2n can be broken down into the form 2n ×odd number.
For example, 6276
This can be written as (2×31)76=276×3176
We can find the last two digits of both the parts separately.
Also, 210=1024
So, 276 can be written as (210 )7×26.
We can easily find the last two digits of this using previous methods. This can be used to find the last two digits of numbers raised to higher powers.
For example: Find the last two digits of 2543.
2543 = (210 )54 × 23
210 ends in 24
⇒2543 = (24)54 × 23
24 raised to an even power ends in 76
⇒2543 =76 × 8
The last two digits of 2543 will be 08.
If we need to multiply 76 with 2n, then we can straightaway write the last two digits of the product as the last two digits of 2n because when 76 is multiplied with 2n the last two digits remain the same as the last two digits of 2n. This will work only for powers of 2,n≥ 2
For example, the last two digits of 76 × 27 will be the last two digits of 27 = 28.
We consolidate the data as given below:
Solved Examples
Qno.1: Find the last two digits of 
.
Solution: We know that 76 raised to any power gives the last two digits as 76. 
Since this is a power of 76, the last two digits will also be 76.
Qno.2: Find the last two digits of 111111.
Solution: Given number is 111111 and here the number is ending at 1.
For a number of the form xy where x ends in 1 the two steps given below are followed to find the last two digits.
Multiply the tens digit of the number x with the last digit of the exponent y to get the tens digit.
Tens digit =1×1=1 The unit digit is equal to one.
Hence, the last two digits of ?111?^111 is 11.
Qno.3: Find the last two digits of 82386.
Solution: The numbers which are not in the form of 2n can be broken down into the form 2n ×odd number.
82386=(2×41)386=2386×41386=2380×26×41386=(210 )38×26×41386
210 ends in 24. 24 raised to an even power ends in 76.
26=64
The tens digits of 41386=1×6=6
Unit digit is 1.
So, the last two digits of 41386=61
So, the last two digits of 82386= Last two digits of the product 76×64×61=04