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

Let's simplify : $2^{122} + 4^{62} + 8^{42} + 4^{64} + 2^{130}$

$= 2^{122} + 2^{2 \times{62}} + 2^{3 \times{42}} + 2^{2 \times {64}} + 2^{130}$

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

$= 2^{122} + 2^{122} \times 2^2 + 2^{122} \times 2^4 + 2^{122} \times 2^6 +2^{122} \times 2^8$

$= 2^{122} (1 + 4 + 16 + 64 + 256)$

$= 2^{122} (341)$

$341$ is divisible by $11$ , hence, entire expression will be divisible by 11.

Hence, the correct answer is Option 4.