--- title: "8086 ALP: sum of 100 even numbers" subject: "Microprocessor" chapter: 3 key: "semester_4/microprocessor:alp-8086-sum-even-numbers" source: https://www.bctengineer.com/q/semester_4/microprocessor/alp-8086-sum-even-numbers site: BCT Engineering --- # 8086 ALP: sum of 100 even numbers Microprocessor · Chapter 3: Programming with 8086 Microprocessor - Asked in: 2074 Bhadra, 8 marks - Video: [Program to find sum of even numbers from 1 to 100 in assembly language | DosBox | 8086 | Sameh UlHaq](https://www.youtube.com/watch?v=_km-xkj9WkM) — Sameh Ulhaq - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/microprocessor/alp-8086-sum-even-numbers - Whole chapter: https://www.bctengineer.com/md/semester_4/microprocessor/chapter-3 ## Question Write an assembly language program for 8086 to calculate the addition of 100 natural even numbers and display the result in screen, in decimal format. ## Short answer Sum the first 100 even numbers (2+4+...+200) in a loop using DX as the current even, CX as counter, AX accumulating sum. The result 10100 is converted to decimal by repeated division by 10, pushing remainders, then popping and displaying as ASCII digits. Output: 10100. ## Answer The first 100 even natural numbers are 2, 4, 6, ..., 200. Their sum = 100 × 101 = 10100, which fits in a 16-bit register. The program adds them in a loop (DX holds the current even number), then converts AX to decimal by repeated division by 10 (remainders are pushed on the stack and popped in reverse order) and displays the digits. ```asm .MODEL SMALL .STACK 100H DISPSTR MACRO MSG ; display a '$'-terminated string LEA DX,MSG MOV AH,09H INT 21H ENDM CLRSCR MACRO ; clear screen and home the cursor MOV AX,0600H ; scroll whole window (AL=0 -> blank it) MOV BH,07H ; normal attribute MOV CX,0000H ; upper-left corner (row 0, col 0) MOV DX,184FH ; lower-right corner (row 24, col 79) INT 10H MOV AH,02H ; set cursor position MOV BH,00H ; page 0 MOV DX,0000H ; row 0, col 0 INT 10H ENDM .DATA MSG DB 'Sum of first 100 even numbers = $' .CODE MAIN PROC MOV AX,@DATA MOV DS,AX XOR AX,AX ; AX = sum = 0 MOV DX,2 ; first even number MOV CX,100 ; 100 numbers ADDLP: ADD AX,DX ; sum = sum + even number ADD DX,2 ; next even number LOOP ADDLP ; repeat 100 times, AX = 10100 PUSH AX ; save the sum DISPSTR MSG POP AX MOV BX,10 ; divisor XOR CX,CX ; digit counter DIVLP: XOR DX,DX ; DX:AX / 10 DIV BX ; AX = quotient, DX = remainder PUSH DX ; save digit INC CX OR AX,AX ; quotient zero? JNZ DIVLP SHOWLP: POP DX ; most significant digit first ADD DL,30H ; convert to ASCII MOV AH,02H INT 21H ; display digit LOOP SHOWLP MOV AH,4CH ; return to DOS INT 21H MAIN ENDP END MAIN ``` **Output:** `Sum of first 100 even numbers = 10100` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the sum of the first 100 even natural numbers and why does it fit in AX? A. 5050, fits in 8 bits B. 10100, fits in 16 bits C. 10100, needs DX:AX because it exceeds 16 bits D. 20100, fits in 16 bits Answer: B. Sum = 100 x 101 = 10100, which is below 65535, so one 16-bit register is enough. (Key point: Sum value) 2. How does the addition loop generate the even numbers? A. DX starts at 2 and ADD DX,2 after each addition, repeated 100 times with LOOP B. DX starts at 0 and is incremented by 1 each time C. CX is added directly to AX D. DX starts at 1 and ADD DX,2 each time Answer: A. DX holds the current even number, starting at 2 and increasing by 2; CX=100 controls the count. (Key point: Loop logic) 3. How is AX converted to decimal digits for display? A. Repeated multiplication by 10 and taking the high byte B. Repeated division by 10, pushing remainders on the stack C. Adding 30H once to the whole of AX D. Repeated subtraction of 100 until zero Answer: B. DIV by 10 gives the remainder as the last digit; remainders are pushed and later popped. (Key point: Decimal conversion) 4. Why are the remainders pushed on the stack before display? A. To save memory B. Because INT 21h only reads from the stack C. Division gives the least significant digit first, and popping reverses it to most significant first D. To convert them to ASCII automatically Answer: C. The stack is LIFO, so the digit produced last (most significant) is popped first. 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