Bits for decimal 1000 – UGC NET Question

We require n such that 2ⁿ − 1 ≥ 1000 for unsigned representation. Checking nearby powers, 2⁹ = 512 and 2¹⁰ = 1024. Since 512 ≤ 1000 ≤ 1023, 9 bits are insufficient, but 10 bits cover up to 1023. Thus, 10 bits are the minimum needed to represent 1000 in binary.

Option A:
Option A correctly applies the inequality 2ⁿ−1 ≥ N with N = 1000. It identifies that 2⁹ − 1 = 511 is too small, while 2¹⁰ − 1 = 1023 is adequate. Therefore, 10 is the least n satisfying the condition, making this option correct.

Option B:
Option B, 9 bits, corresponds to a maximum value of 511, which does not reach 1000. Using only 9 bits would fail to encode the number 1000, so it cannot be minimal or sufficient.

Option C:
Option C, 11 bits, is technically sufficient but not minimal, since 10 bits already handle values up to 1023. The question specifically asks for the minimum, so including an extra bit is unnecessary.

Option D:
Option D, 8 bits, yields a maximum of 255, which is far below 1000. It clearly cannot represent 1000 and hence is not even sufficient, let alone minimal.