2^{16}-1 is divisible by | vedclass

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(C) We can write $2^{16}-1$ as a difference of squares: $2^{16}-1 = (2^8)^2 - 1^2$ Using the identity $a^2 - b^2 = (a-b)(a+b)$,we get: $(2^8 - 1)(2^8

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) We can write $2^{16}-1$ as a difference of squares: $2^{16}-1 = (2^8)^2 - 1^2$ Using the identity $a^2 - b^2 = (a-b)(a+b)$,we get: $(2^8 - 1)(2^8 + 1)$ Since $2^8 = 256$,we have: $(256 - 1)(256 + 1) = 255 	imes 257$ Now,we check for divisibility by the given options: $255 = 15 	imes 17$. Since $255$ is a factor of $2^{16}-1$,the expression is divisible by $17$.

$2^{16}-1$ is divisible by

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