If the numerical value of a 2-byte unsigned integer on a little endian computer is 255 more than that on a big endian computer, which of the following choices represent(s) the unsigned integer on a little endian computer?

Let the two consecutive memory bytes be $B_0$ (at the lower memory address) and $B_1$ (at the higher memory address).

• In a big-endian system, the most significant byte (MSB) is stored at the lower memory address, so the numerical value is:

  $$V_{\text{BE}} = 256 \cdot B_0 + B_1$$

• In a little-endian system, the least significant byte (LSB) is stored at the lower memory address (and MSB at the higher address), so the numerical value is:

  $$V_{\text{LE}} = 256 \cdot B_1 + B_0$$

Given that $V_{\text{LE}} - V_{\text{BE}} = 255$:

Thus, $B_1 = B_0 + 1$.

The representation of the unsigned integer value on the little-endian computer is written with MSB followed by LSB, i.e., byte $B_1$ followed by byte $B_0$.

Let us evaluate each option $(B_1B_0)_{16}$:

• (A) 0x6665: $B_1 = 0\text{x}66$, $B_0 = 0\text{x}65 \implies B_1 - B_0 = 0\text{x}66 - 0\text{x}65 = 1$. (Correct)

• (B) 0x0001: $B_1 = 0\text{x}00$, $B_0 = 0\text{x}01 \implies B_1 - B_0 = -1 \neq 1$. (Incorrect)

• (C) 0x4243: $B_1 = 0\text{x}42$, $B_0 = 0\text{x}43 \implies B_1 - B_0 = -1 \neq 1$. (Incorrect)

• (D) 0x0100: $B_1 = 0\text{x}01$, $B_0 = 0\text{x}00 \implies B_1 - B_0 = 0\text{x}01 - 0\text{x}00 = 1$. (Correct)

Hence, the correct choices are (A) and (D).