What is the highest decimal value that can be represented using 10 binary digits (10 bits)? Explain the 2^n − 1 rule.

Introduction / Context:
Bit width determines numeric range. For unsigned binary numbers, the maximum representable value with n bits is 2^n − 1 because all bits set to 1 yields that sum. Understanding this rule is critical for designing counters, address buses, and buffer size limits.

Given Data / Assumptions:

• The quantity is an unsigned value using exactly 10 bits.
• Each bit represents a power of two: 2^0 through 2^9.
• We seek the largest value with all bits set to 1.

Concept / Approach:
For n bits, the range is 0 to (2^n − 1). With 10 bits, that is 0 to (2^10 − 1). Since 2^10 = 1024, the highest value is 1023. This result follows from the geometric series sum: 1 + 2 + 4 + … + 2^(n−1) = 2^n − 1, which is exactly the value when all n bits are 1s.

Step-by-Step Solution:
Compute 2^10: 1024.Apply the formula for the maximum unsigned value: 2^n − 1 → 1024 − 1.Obtain 1023 as the highest representable value.Select option “1023.”

Verification / Alternative check:
Write the 10-bit all-ones pattern: 1111111111 (binary). Converting to decimal: 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 1023, confirming the formula-based result.

Why Other Options Are Wrong:
1024 equals 2^10 and requires 11 bits to represent as an unsigned value (1 followed by ten zeros). 512 is 2^9 and is far below the maximum. “All of the above” cannot be correct because only one value is the true maximum.

Common Pitfalls:
Confusing the count of distinct values (2^n) with the maximum value (2^n − 1); forgetting that the smallest value is 0, not 1, for unsigned ranges.

Final Answer:
1023