$$2^{122}+4^{62}+8^{42}+4^{64}+2^{130}$$ is divisible by which one of the following integers?

The given expression is $$2^{122}+4^{62}+8^{42}+4^{64}+2^{130}$$

We can also write this expression as: $$2^{122}+\left(2^2\right)^{62}+\left(2^3\right)^{42}+\left(2^2\right)^{64}+2^{130}$$

=$$2^{122}+2^{124}+2^{126}+2^{128}+2^{130}$$

=$$2^{122}\left(1+2^2+2^4+2^6+2^8\right)$$

=$$2^{122}\left(\dfrac{1\times\ \left(2^2\right)^5-1}{2^2-1}\right)$$

=$$2^{122}\left(\dfrac{2^{10}-1}{2^2-1}\right)$$

=$$2^{122}\left(\dfrac{1023}{3}\right)$$

=$$2^{122}\times\ 341$$

Now, 341=31*11, so it is divisible by 11

So, correct answer is option D.