Challenge Description#

Given a 16-bit DOS assembly program. Inputs are:

• input1 = 20
• input2 = 200

Flag format:

scsc26{mem[001E]_mem[001F]_DX_BX}

Analysis#

At a high level, the program:

• Reads 2 inputs using DOS int 21h function AH=0Ah (buffered input).
• Copies input into number3 and number4 from the back (uses STD + LODSB/STOSB) to right-align digits (4-digit style).
• Treats filler bytes as empty, effectively as 0.
• Takes 4 digits from each number, converts ASCII '0'..'9' to numeric 0..9.
• Adds digit-by-digit with carry and stores a 5-digit result into memory 0x001C..0x0020.

With right-alignment:

• 20 becomes 0020
• 200 becomes 0200
• Sum = 00220

Solution#

Because the result is stored as digits:

• mem[001E] (hundreds digit) = 02
• mem[001F] (tens digit) = 02

At the end of the loop:

• DX = 1c00 (DH=1C, DL=00)
• BX = 001c

So the flag is:

scsc26{02_02_1c00_001c}