As you are preparing for aptitude I would prefer giving you some short cuts to find last 2 digits which will save you time
$(odd\ number)^{power}$
the last 2 digits only depends on last digit of power and the last 2 digits of the odd number given
CASE 1: if the odd number ends in 1 like $(y1)^{n}$ last 2 digits are (units digit of y*n)(1)
if odd number ends in 7 , 3 , 9 trick is first bring it to a form that ends in 1.
keep in mind that $7^4$ , $9^2$ , $3^4$ ends in 1
CASE 2: if the number is like $(x7)^{4}$ then last two digits are (units digit of 2*x)(1)
CASE 3: if the odd number is like $(x3)^{4}$ then last two digits are (8 - [units digit of 2*x])(1)
CASE 4 : if the odd number is like $(x9)^{2}$ then last two digits are (8 - [units digit of 2*x])(1)
now having made numbers ending in 1 , then the process is as in case 1 with remaining of power after removing 4 for case 2 and case 3 or 2 for case 4
Now $(2)^{power}$
shortcut is as follows
CASE 5 : if ten's place of power is even the last 2 digits are just last 2 digits of $2^{last\ digit\ of\ power}$
CASE 6 : if ten's place of power is odd the last 2 digits are just last 2 digits of $3 \times 2^{last\ digit\ of\ power + 3}$
so case 1,2,3,4 is for number ending in odd number
case 5,6 for power of 2
as any even number can be expressed as multiple of 2 and an odd number a combination ,of one of first 4 cases and one of case 5 or case 6 will fetch you the answer for $(even\ number ) ^{power}$
with the above short cut techniques I am able to calculate last 2 digits in a matter of 15 to 20 seconds without calculator