Difficulty: Medium
Correct Answer: 10
Explanation:
Introduction / Context:
Determining the required bit width for a counter, address space, or identifier is an everyday design decision. Using too few bits causes overflow; too many wastes resources. The key is relating the count of representable states to powers of two.
Given Data / Assumptions:
Concept / Approach:
Find successive powers of two until the count meets or exceeds the required number of states. This is equivalent to computing the ceiling of log2(748).
Step-by-Step Solution:
Compute nearby powers: 2^8 = 256, 2^9 = 512, 2^10 = 1024.Compare to requirement: 512 < 748 ≤ 1024.Therefore, 9 bits are insufficient, but 10 bits suffice.Minimum bit-width = 10 bits.
Verification / Alternative check:
Using logarithms: log2(748) ≈ ln(748)/ln(2) ≈ 6.619/0.693 ≈ 9.55. Taking the ceiling gives 10. This matches the power-of-two comparison method.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
10
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