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Official PYQ · 1995

GATE CS 1995 Q2.12 — Number Representations (Fixed and Floating Point)

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The number of 11's in the binary representation of (3∗4096+15∗256+5∗16+3)(3\ast4096 + 15\ast256 + 5\ast16 + 3) are:

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Answer: C

We can find the number of 11s efficiently by converting the given expression into its hexadecimal (base-16) representation.

Expressing the given terms as powers of 1616:

  • 4096=1634096 = 16^3
  • 256=162256 = 16^2
  • 16=16116 = 16^1
  • 1=1601 = 16^0

Substituting these into the expression:

N=3×163+15×162+5×161+3×160N = 3 \times 16^3 + 15 \times 16^2 + 5 \times 16^1 + 3 \times 16^0

In hexadecimal, this is represented as (3F53)16(3\text{F}53)_{16} (since 1515 corresponds to F\text{F} in hex).

Now, convert each hexadecimal digit to its 4-bit binary equivalent and count the number of 11s:

  • 316=001123_{16} = 0011_2 → contains 2 ones
  • F16=11112\text{F}_{16} = 1111_2 → contains 4 ones
  • 516=010125_{16} = 0101_2 → contains 2 ones
  • 316=001123_{16} = 0011_2 → contains 2 ones

Total number of 11s = 2+4+2+2=102 + 4 + 2 + 2 = 10.

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