Bits needed up to one million – UGC NET Question

To represent all integers from 0 to 10⁶ inclusive, we need at least 10⁶ + 1 distinct bit patterns. In binary, n bits can represent 2ⁿ distinct values, so we require 2ⁿ ≥ 10⁶ + 1. Since 2¹⁹ = 524288 < 10⁶ + 1 and 2²⁰ = 1048576 ≥ 10⁶ + 1, we must take n = 20. Therefore, 20 bits are the minimum needed, making option C correct. Option A: Option A, 19 bits, can represent only 524288 values, fewer than the 1,000,001 needed (0 through 10⁶). So it is insufficient. Option B: Option B, 18 bits, represents an even smaller range, so it cannot work. Option C: Option C is the smallest n with 2ⁿ ≥ 1,000,001, so it is minimal and correct. Option D: Option D, 21 bits, is sufficient but not minimal since 20 bits already cover the required range.