Address lines and addressable locations in digital memory If a memory device provides 12 address lines (A0 through A11), how many unique storage locations can be addressed in total?

Introduction / Context:
In digital memory systems, the number of address lines determines how many unique locations can be accessed. Each distinct binary combination on the address bus selects exactly one storage location. Understanding this power-of-two relationship is foundational for memory sizing and address decoding.

Given Data / Assumptions:

• Number of address lines n = 12 (A0–A11).
• Each unique address corresponds to one storage location (regardless of data width).
• Binary addressing yields combinations equal to powers of two.

Concept / Approach:

The total number of unique addresses provided by n binary lines is 2^n. This comes from combinatorics: each line can be 0 or 1, and the combinations multiply (2 * 2 * … * 2, n times). The data bus width (for example, 8 bits or 16 bits) affects bits per location, not the count of locations.

Step-by-Step Solution:

Verification / Alternative check:

Quick mental check using nearby powers: 2^8 = 256, 2^16 = 65,536. Twelve is midway between, giving 4096, which is consistent with standard memory sizing (4K locations).

Why Other Options Are Wrong:

• 144: not a power of two; does not correspond to any integer number of address lines.
• 512: equals 2^9 (would require 9 address lines, not 12).
• 2048: equals 2^11 (would require 11 address lines).

Common Pitfalls:

• Confusing number of locations with total bits (which also depends on data width per location).
• Assuming decimal growth rather than exponential power-of-two growth with additional lines.

Final Answer:

4096