Storage sizing — bytes needed for a value: Determine the minimum whole bytes required to store the unsigned decimal value 2875 in binary (no sign bit, no leading zeros).

Introduction / Context:Memory and register sizing problems are routine in digital design. The key steps are finding the minimum number of bits required for a given unsigned value, then converting that bit count to whole bytes (8-bit groups). Here we size storage for the decimal value 2875.

Given Data / Assumptions:

• Unsigned value: 2875.
• No sign bit; fixed binary storage.
• 1 byte = 8 bits; we must use a whole number of bytes.

Concept / Approach:Compute n = floor(log2(N)) + 1 to get the minimal bit length, or bound N between powers of two. Then compute bytes = ceil(n / 8). This ensures there is enough capacity without wasted fractional bytes.

Step-by-Step Solution:Find surrounding powers of two: 2^11 = 2048, 2^12 = 4096.Since 2048 ≤ 2875 ≤ 4095, the value requires 12 bits.Translate to bytes: ceil(12 / 8) = ceil(1.5) = 2 bytes.Therefore, two bytes suffice to store 2875 in binary.

Verification / Alternative check:Direct conversion to binary (optional) will produce a 12-bit pattern, reaffirming that 16-bit (2-byte) storage is the nearest whole-byte capacity.

Why Other Options Are Wrong:3, 4, 5: These provide more space than needed. 3 bytes would cover up to 24 bits (overkill for a 12-bit value).

Common Pitfalls:Using 8 bits per byte incorrectly or adding an unnecessary sign bit; forgetting to round up to the next whole byte after computing the bit length.

Final Answer:2