KVPY 2020-21 SB SX Exam Previous Papers

Question : 18 of 120

Marks: +1, -0

Let

{M}=2^{30}-2^{15}+1

, and

{M}^{2}

be expressed inbase 2 . The number of l's in this base 2 representation of

{M}^{2}

Solution:

(2^{{n}})_{2}=1 {00 \....0}_{{n} }

{M}^{2}=(2^{60}-2^{46})+(2^{30}-2^{16})+2^{31}+1

=({\{11 \..... 1}\} \{{00 \..... 0}\}+{\{11 \.....1}\}

\{{000 \......0} \}+\1{00 \.....0\}+1)_2

Number of 1's = 14 + 1 + 14 + 1 = 30

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