Binary length estimation — how many bits does a decimal need? In base-2 representation, determine how many binary digits (bits) are required to represent the decimal number 93 without leading zeros.

Introduction / Context:
Knowing how many bits are required to represent a given unsigned decimal number in binary is a foundational skill in digital electronics, impacting memory sizing, register widths, and bus design. Here, we determine the minimum number of binary digits needed to represent the decimal value 93 without using leading zeros.

Given Data / Assumptions:

• Unsigned representation (no sign bit).
• Value: 93 in base-10.
• No leading zeros; we want the minimal bit length.

Concept / Approach:
The number of bits n required to represent a positive integer N satisfies 2^(n-1) ≤ N ≤ 2^n - 1. Equivalently, n = floor(log2(N)) + 1. Alternatively, we can convert 93 to binary and count the digits directly, or bound it between nearby powers of two.

Step-by-Step Solution:
Compute powers of two near 93: 2^6 = 64 and 2^7 = 128.Since 64 ≤ 93 ≤ 127, any binary form of 93 must fit within 7 bits (range for 7 bits is 0 to 127).Optional explicit conversion: 93 = 64 + 16 + 8 + 4 + 1 → binary 1011101.Count digits in 1011101 → 7 bits.

Verification / Alternative check:
Using n = floor(log2(93)) + 1 → log2(93) is between 6 and 7, so floor(·) = 6 → n = 7. This matches the direct conversion count.

Why Other Options Are Wrong:
eight: 8 bits can represent 93 but are not required; the question asks for the minimal number.
11: Far too many for such a small value; 11 bits cover up to 2047.
five: 5 bits cover only up to 31; insufficient for 93.

Common Pitfalls:
Confusing the maximum representable value with the minimum needed bits, or accidentally adding a sign bit when the question specifies an unsigned value.

Final Answer:
seven