A question related to division

Question

$2^{21} + 1024^{2} + 16^6$ is divisible by: a) 31 b) 19 c) 13 d) 17

My attempt: I tried using the remainder theorem and the factor theorem. But, nothing. Pls help me in moving forward

Answer 1

Hints:
$1024 ^{2} = (2^{10})^2 = 2^{20}$

$16^6 = (2^4)^6 = 2^{24}$

So you have $2^{21} + 2^{20} + 2^{24}$

Can you do now?

Answer 2

First of all simplify the second term and third term.

$1024 ^{2} = (2^{10})^2 = 2^{20}$

$16^6 = (2^4)^6 = 2^{24}$

So now we have $ 2^{21} + 2^{20} + 2^{24} $

Simplifying it further $ 2^{20}(2 + 1 + 16) $

$ 2^{20}(19) $

Therefore it is divisible by 19