Digital memory addressing: If a memory device provides twelve (12) address lines from the processor or address decoder, how many unique storage locations (addresses) can it select in total?

Introduction / Context:
Address lines determine how many distinct memory locations a digital system can reference. This question checks your understanding of the binary addressing capacity derived from the number of address lines, a foundational concept in microprocessor and memory design.

Given Data / Assumptions:

• Number of address lines: N = 12.
• Each address line carries a binary value (0 or 1).
• We assume a flat, linearly addressed memory space without bank switching.

Concept / Approach:
With N independent binary lines, the number of unique combinations is 2^N. Each unique combination selects one unique address (storage location). Therefore, the capacity in addresses equals 2 raised to the power of the number of address lines.

Step-by-Step Solution:
Let N = 12 address lines.Compute number of addresses: total_addresses = 2^N.Evaluate: 2^12 = 4096.Therefore, the memory can address 4096 unique storage locations.

Verification / Alternative check:
Build up powers of two: 2^10 = 1024, then multiply by 4 (2^2) to reach 2^12 → 1024 * 4 = 4096. This mental math provides a fast confirmatory check.

Why Other Options Are Wrong:
144: not a power of two; address counts are always powers of two.512: equals 2^9, which would correspond to only 9 address lines.2048: equals 2^11, which would require 11 address lines, not 12.

Common Pitfalls:
Confusing bytes with addresses. Address count depends solely on address lines. Byte/word organization pertains to data bus width and memory organization, not the raw count of addresses.

Final Answer:
4096