Discussion
For a memory unit of 4096 words, how many bits are needed to specify the address?
Answer: B
A memory of \(N\) words needs \(n\) address bits with \(2^n \ge N\).
\[ 4096 = 2^{12} \]
So 12 address bits select any of the 4096 words. Eight bits would address only 256 words and 15 bits would address 32768 words, more than needed, and 4 bits only 16. The answer is 12, option B.
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