Question

It is being given that 2^32 is completely divisible by a whole number. Which of the following numbers is completely divisible by this number? (A) 216+ 1 (C) 7 x 223 (B) 216-1 (D) 296+1​

Answer 1

Answer with explanation:

Factors of, $2^{32}=1,2,2^2,2^3,2^4,2^5,2^6,2^7,2^8,2^9,..........2^{32}$

So, all of them are whole number which will divide $2^{32}$ completely.

Now, coming to the options

Apart from ,$7 \times2^{23},$ none of the three ,$2^{16}+1, 2^{16}-1,2^{96}+1$, are factor of , $2^{32}$.

So, $7 \times2^{23},$ will divide all factors of $2^{32}$, which have exponent of 2 greater than or equal to 23.

$7 \times2^{23},$ is completely divisible by this number.

Answer 2

Answer:

not correct poda pulla