Write an assembly language program that corresponds to the following C program: int num; int main () { scanf("%d", &num); num = num % 16; printf("num = %d\n", num); return 0; }

Convert the following to Pep/9 assembly: #include <iostream>using namespace std;int main(){  int number;  cin >> number;  if (number % 2 == 0) {     cout << “Even\n”;  }  else {    cout << “Odd\n”;  }  return 0;}

Define the term pointer variables.

15. Translate the following C program to Pep/9 assembly language. #include int main() { int numitms, j, data, sum; scanf("%d", &numItms); sum = 0; for (j = 1; j <= numitms; j++) { scanf("%d", &data); sum += data; } printf ("Sum: %d\n", sum); return 0; } Sample Input 48-376 Sample Output Sum: 18

please modify the c language program into a pep 9 assembly language  #include <stdio.h> int num, answer; int divide ( int numer, int denom ) { int quotient, remain; remain = numer; quotient = 0; while ( remain >= denom ) { remain -= denom; quotient++; } // end for return quotient; } // end of divide( ) int modulus ( int n, int d ) { int quot, rem; rem = n; quot = 0; while ( rem >= d ) { rem -= d; quot++; } // end for return rem; } // end of modulus( ) int main() { printf("? "); scanf("%d", &num); answer = divide(num, 20); printf("%d / 20 = %d\n", num, answer); answer = modulus(num, 20); printf("%d mod 20 = %d\n", num, answer); answer = divide(num, 35); printf("%d / 35 = %d\n", num, answer); answer = modulus(num, 35); printf("%d mod 35 = %d\n", num, answer); return 0; } // end of main