Finding the last two digits of large numbers, written in an indirect exponential form, is a challenging task in quantitative aptitude problems. For example, to find the last two digits of 2^11 by continuous multiplication is reasonably less time consuming but finding the last two digits of2^110 is not. To find the last two digits of large numbers, we divide numbers into cases.
Numbers ending with 1
The units digit, as discussed in the corresponding article is 1. The tens digit is the units digit of the product of the index and the tens digit of the base, i.e a number abc1^xy will have the last digit as 1 and the second last digit or the tens digit equal to the units digit of the product y x c.
For example, 441^67 will end with 81.
Numbers ending with 3, 7 or 9
As shown in the units digit article, 3^4, 3^8, 3^12... 3^4k where k = 0, 1, 2...and so on, end in 1. Similarly all values of 7^4k and 9^2k where k = 0, 1, 2, 3... and so on end in 1. Therefore, to calculate the last two digits of numbers ending with 3, 7 or 9 convert them to a form in which they end in 1 and calculate.
For example, 443^61 can be rewritten as (443^4)^15 x 443. 443^2 will end in 49 and 49^2 will end in 01. Therefore 443^4 will end in 01. 443^60 will end in 01. Therefore 443^61 will end in 01 x 43 = 43.
Numbers ending with 2, 4, 6, 8
It is an empirical observation that 76 when raised to any power, ends in 76. Also 24^2 ends in 76. Also, we know that 2^10 is 1024. Therefore, 1024, when raised to any even power, will give 76 and when raised to an odd power will give 24. This fact helps to find the last two digits of any even number. Further all even numbers can be written as 2^(power) x (odd number). The method to find the last two digits of an odd number have already been discussed above.
For example, 56^283 = (2^3 x 7)^283 = 2^ 849 x 7^283
= (2^10)^84 x 2^9 x (7^4)^70 x 7^3
=76 x 12 x 01 x 43 = 16