​$$4^{38}+4^{40}-4^{39}+4^{41}$$​ is divisible by which one of the following?

Correct option is A

Given:

​$$4^{38}+4^{40}-4^{39}+4^{41}$$

Solution:

$$4^{38}+4^{40}-4^{39}+4^{41}$$​

= $$4^{38}[1+4^2-4^1+4^3]$$

=$$4^{38}[1+16-4+64]$$

=$$4^{38}\times 77$$

=$$4^{38}\times 7\times 11$$

Hence, $$4^{38}+4^{40}-4^{39}+4^{41}$$ is divisible by 7.​