Let M = 2^{30} - 2^{15} + 1. When M^2 is expr | vedclass

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(B) Given $M = 2^{30} - 2^{15} + 1$. We know that $(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc$. Let $a = 2^{30}$,$b = 2^{15}$,and $c = 1$. Then

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$30$

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(B) Given $M = 2^{30} - 2^{15} + 1$. We know that $(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc$. Let $a = 2^{30}$,$b = 2^{15}$,and $c = 1$. Then $M^2 = (2^{30})^2 + (2^{15})^2 + 1^2 - 2(2^{30})(2^{15}) + 2(2^{30})(1) - 2(2^{15})(1)$. $M^2 = 2^{60} + 2^{30} + 1 - 2^{46} + 2^{31} - 2^{16}$. $M^2 = 2^{60} - 2^{46} + 2^{31} + 2^{30} - 2^{16} + 1$. $M^2 = 2^{60} - (2^{46} - 2^{31}) + 2^{30} - 2^{16} + 1$. $M^2 = 2^{60} - 2^{31}(2^{15} - 1) + 2^{30} - 2^{16} + 1$. Alternatively,using $M = 2^{15}(2^{15} - 1) + 1$,we observe the pattern of binary digits. Expanding $M^2 = (2^{30} - 2^{15} + 1)^2 = 2^{60} - 2^{46} + 2^{31} + 2^{30} - 2^{16} + 1$. $M^2 = 2^{60} - (2^{46} - 2^{31}) + 2^{30} - 2^{16} + 1$. $M^2 = 2^{60} - (2^{45} + 2^{44} + \dots + 2^{31}) + 2^{30} - 2^{16} + 1$. After simplification,the binary representation contains $30$ ones.

Let $M = 2^{30} - 2^{15} + 1$. When $M^2$ is expressed in base $2$,the number of $1$'s in its binary representation is:

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