Question

Find the last two digits of the expansion (212n-64n) when n is a positive integer?

a. 5

b. 11

c. 10

d. 00

Answer 1

Following are the options for this question:

The correct answer to this question is

(B) 00.  

Substitute n = 1.  

2^ 12 - 6^4 = 4096 - 36^2 = 4096 - 1296 = 2800.

Solution 2:

2^(12n) - 6^(4n)  

(2^6n)^2 - (6^2n)^2 ===> (64^n + 36^n)(64^n-36^n)  

===> first term will lead to last 2 digits of 00 for all odd n, and second will be 00 for all even n, so ans is "B"