If $$N = 4^{11} + 4^{12} + 4^{13} + 4^{14}$$, then how many positive factors of $$N$$ are there?

$$N=4^{11}+4^{12}+4^{13}+4^{14}$$

or, $$N=4^{11}\left(1+4^1+4^2+4^3\right)=4^{11}\times5\times17\ .$$

So, $$N=2^{22}\times5\times17\ .$$

Case1 (1 factor) :

$$2,2^2,2^3,....,2^{22},5,17\ $$

Total 24 factors .

Case2 (2 factors) :

$$2\times5,2\times17,\ 2^2\times5,2^2\times17,....,5\times17\ .$$

Total $$2\times22+1=45\ $$factors .

Case3 (3 factors) :

$$2\times5\times17,\ 2^2\times5\times17,.......,2^{22}\times5\times17\ .$$

Total $$22$$ factors .

Case4 :

1 is also a factor .

Total number of factors = $$24+45+22+1=92\ .$$

A is correct choice.