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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(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:

Similar Questions

$\sum_{\substack{i, j=0 \\ i \neq j}}^{n} {}^{n}C_{i} {}^{n}C_{j}$ is equal to

Find the coefficient of $a^{4}$ in the product $(1+2a)^{4}(2-a)^{5}$ using the binomial theorem.

Let $(1+x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$. If $(a_1+a_3+a_5+\ldots+a_{19})-11 a_2=121 k$,then $k$ is equal to . . . . . . .

The value of $\sum\limits_{n = 0}^4 {{{\left( {1009 - 2n} \right)}^4} \binom{4}{n} {\left( { - 1} \right)^n}}$ is

Let $R=(5 \sqrt{5}+11)^{2 n+1}$ and $f=R-[R]$,where $[x]$ denotes the greatest integer less than or equal to $x$,then $R f=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module