A microcomputer with 64K of addressable memory has hexadecimal addresses ranging from the first location 0000 to the last location FFFF (inclusive). Identify the correct first and last addresses.

Introduction / Context:
Address ranges in microcomputers are often expressed in hexadecimal. A “64K” memory space contains 65,536 addresses, typically numbered from 0000 to FFFF in hex, where each address points to one byte. Recognizing these limits is foundational for memory maps, stack placement, and I/O decoding.

Given Data / Assumptions:

• Total size is 64K bytes = 65,536 decimal addresses.
• Hexadecimal digits span 0–F, so four hex digits represent 0–65,535.
• Addressing is byte-addressable and contiguous.

Concept / Approach:
Four hex digits cover 16^4 = 65,536 values. The first value is 0000 (decimal 0). The last value is FFFF (decimal 65,535). Therefore, a 64K space exactly fits within 0000–FFFF. Other pairs either undercount or are not valid hex ranges for 64K.

Step-by-Step Solution:
Compute range size: 16^4 = 65,536 addresses.Map first address: 0000.Map last address: FFFF.Select the pair that matches this mapping.

Verification / Alternative check:
Decimal check: 0000 = 0, FFFF = 65,535; count of addresses = 65,536 = 64 * 1024 = 64K, confirming correctness.

Why Other Options Are Wrong:
0000, EEEE: ends at 61,182 decimal, not 64K.0, 64: mixes decimal and lacks 4-digit hex formatting.0000, 9999: stops at 39,321 decimal; not the 64K boundary.None of the above: incorrect since 0000, FFFF is correct.

Common Pitfalls:
Confusing K as 1000 vs 1024; forgetting that hex FFFF equals 65,535; mixing numeric bases when reading address ranges.

Final Answer:
0000, FFFF