Jumper settings and binary – Highest number with a 3-position jumper block Assuming a three-position jumper block represents three binary bits (each position can be 0 or 1), what is the highest binary number that can be set?

Introduction / Context:
Hardware options such as drive IDs or feature enables are often configured using jumper blocks. Each jumper position can represent a binary bit, and understanding the numeric range prevents misconfiguration.

Given Data / Assumptions:

• Three independent jumper positions acting like three bits.
• Each bit can be 0 (open) or 1 (jumpered/closed) according to the device's convention.
• No tri-state or resistor ladder—simple binary on/off.

Concept / Approach:

With three bits, the count of unique values is 2^3 = 8, ranging from 000 to 111 in binary. The highest value is 111, which equals 7 in decimal (and 7 in hex). None of the offered numeric choices (4, 6, F, 1) equals 7, so the correct selection must be “None of the above.”

Step-by-Step Solution:

Verification / Alternative check:

List all combinations: 000(0), 001(1), 010(2), 011(3), 100(4), 101(5), 110(6), 111(7). The largest is 7, confirming the conclusion.

Why Other Options Are Wrong:

4 and 6 are valid states but not the maximum; 1 is far below maximum; F corresponds to 1111 (four bits) and is impossible with only three jumpers.

Common Pitfalls:

Assuming hexadecimal naming without enough bits; confusing “three-position jumper block” with a 3-pin single jumper that selects among two states (that scenario uses one bit, not three independent bits).

Final Answer:

None of the above.