The number of 1 's in the binary representation of (3* 4096+15 * 256+5*16+3) are
Write each part in hexadecimal:
\(3 \times 4096 = 3 \times 16^3\), \(15 \times 256 = F \times 16^2\), \(5 \times 16\), \(3\), so the number is \(3F53_{16}\).
In binary: 3 = 0011, F = 1111, 5 = 0101, 3 = 0011, giving 0011 1111 0101 0011.
Number of 1's = 2 + 4 + 2 + 2 = 10.
Answer: (C) 10
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