PROGRAM CODE AND CALCULATION
1
ANSWER 1:-
MSG DB “1” (Message database 1) // press a key to continue 0
NMSG DB “0” (Message database 0)
NUM DB 75H; (Enter the required no 75H means binary 1110101)
START: MOV AX,@DATA (16 bit data register moved in AX)
MOV DS, AX (Move the data register in DS)
MOV AL, NUM (Copy a number into the register)
MOV BL, 02H; (The Separating starts from 2, Therefore BH is link to 02H)
MOV DX,0000H ; (To evade Divide overflow mistake)//DX
MOV AH,00H ; (To evade Divide overflow error)//AH
L1:DIV BL (Loop checking for prime number)
CMP AH,00H ; (Remains is equated with 00H (AH))
JNE NEXT (Conditional Obstacle for the next)
INC BH ; (BH is incremented if the Number is divisible by BL value)
NEXT:CMP BH,02H ; ( compare the value If BH > 02H so it is not prime )
JE FALSE ; ( Jump in equal if statement is not a Prime No)
INC BL ; ( Increase BL value)
MOV AX,0000H ; //(To evade Divide overflow mistake)
MOV DX,0000H ; //(To evade Divide overflow mistake)
MOV AL,NUM ; (Move the Defaulting no to AL)
CMP BL,NUM ; ( Route the loop until BL matches Value).
JNE L1; (Jump to check again with incremented rate of BL number)
//To show the assumed no is a Prime No//
TRUE: LEA DX, MSG (IF statement is true then LEA assign in the DX)
MOV AH, 07H; (Moved the number and Used to print a string)
INT 21H (Assume one integer value)
JMP EXIT ( Jump the exit value and show this number prime or
not)
FALSE: LEA DX, NMSG (If the prime condition is not match then address the in
bit message 0). (Elarde)
MOV AH, 07H; (Used to print a string)
INT 21H (Show this integer value)

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