Let M=2^{30}-2^{15}+1, and M^2 be expressed i | vedclass

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Question

(b)

Given,

$\begin{aligned} M^2=& 2^{66}-2^{46}+2^{32}+2^{30}-2^{16}+1 \ M^2=& 2^{46}\left[\frac{2^{14}-1}{2-1}ight]+2^{32}+2^{16}\le

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(b)

Given,

$\begin{aligned} M^2=& 2^{66}-2^{46}+2^{32}+2^{30}-2^{16}+1 \ M^2=& 2^{46}\left[\frac{2^{14}-1}{2-1}ight]+2^{32}+2^{16}\left[\frac{2^{14}-1}{2-1}ight]+1 \ M^2=& 2^{46}\left[1+2+2^2+\ldots+2^{13}ight] \ & \quad+2^{32}+2^{16}\left[1+2+2^2+\ldots+2^{13}ight]+1 \ M^2=& \frac{2^{59}+2^{58}+\ldots+2^{46}}{14 	ext { terms }}+2^{32} \end{aligned}$

$\underbrace{+2^{29}+2^{28}+\ldots+2^{16}}_{14 	ext { terms }}+2^0$

Therefore, on base $2$ representation of $M^2$, there will be $30$ times digit $1$ .

Let $M=2^{30}-2^{15}+1$, and $M^2$ be expressed in base $2$.The number of $1$'s in this base $2$ representation of $M^2$ is

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