The minimum number of bits required to store any three-digit decimal number is:

Concept

An unsigned binary field with b bits has 2b distinct patterns and represents values from 0 through 2b − 1.

Therefore, the minimum width for a largest required value M is the smallest integer b for which 2b − 1 ≥ M.

Application

A three-digit decimal number can be as large as 999, so the storage width must include 999.

1. With 9 bits, the largest unsigned value is 29 − 1 = 511; the range ends before 999.

2. With 10 bits, the largest unsigned value is 210 − 1 = 1023; the range includes 999.

Cross-check

Equivalently, 29 = 512 < 1000 ≤ 1024 = 210, so ⌈log2(1000)⌉ = 10.

Result

Thus the minimum required width is 10 bits.