Question

Write an assembly language program that receives a number n from input, and determines if n is prime in the following very simple way. After receiving n, the program will determine the remainder of n - 2, n - 3, n - 4, . , testing each possible divisor in turn. You will test all possible divisors, not just the prime ones. (This is not a very efficient program.) As soon as the program finds a number q > 2 such that n - q has remainder 0, the program should send q to output and halt.

Answer 1

The assembly language program determines if a number n is prime by trying divisors from 2 to n-1. The program outputs the first divisor it finds, if any, indicating n is not prime; otherwise, n is prime.

Step-by-step explanation:

The student has asked for an assembly language program that checks if a number n is prime by checking the remainder of n minus every number from 2 up to n - 1. This program stops as soon as it finds a divisor with a zero remainder. Because developing an assembly program depends on the specific assembly language and processor architecture, the following pseudo-assembly code provides a general structure for the task: READ n MOV Q, 2 WHILE Q < n MOV R, n MOD Q IF R == 0 WRITE Q HALT ENDIF INC Q ENDWHILE WRITE 'n is prime'This pseudo-code starts by receiving input n, initializes Q to 2, and enters a loop to check if n modulo Q is zero. If it finds such Q, Q is outputted indicating n is not prime. If the loop finishes without finding a divisor, n is prime.

To determine if a number is prime, we can use a simple assembly language program that tests all possible divisors. First, we will receive the number n from the input. Then, we will subtract 2 from n and determine the remainder. If the remainder is 0, we will output 2 as the divisor and halt. Otherwise, we will subtract 1 from n and determine the remainder again. This process will continue until we find a divisor or until all possible divisors have been tested.Here is an example of how the program could be written in assembly language:Receive the number n from inputSet a counter q to 2Subtract q from n and determine the remainderIf the remainder is 0, output q as the divisor and haltIncrease q by 1 and go back to step 3