Question

make a code for the additions of fractions. use the attached code to find LCM.

#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> int Icm(int, int); int main() int n1, n2, min Multiple; printf("Enter two positive integers: "); scanf("%d %d", &n1, &n2); // maximum number between n1 and n2 is stored in min Multiple int result = Icm(n1, n2); printf("The LCM of %d and %d is %d.\n", n1, n2, result); return 0; int Icm(int n1, int n2){ int min Multiple = (n1 > n2) ? n1 : n2; Il Always true while (1) if (min Multiple % n1 == 0 && minMultiple % n2 == 0) return min Multiple; ++minMultiple;

Answer 1

Program Code to Copy

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

int lcm(int, int);
int gcd(int, int);

int main(){

int num1, num2, num3, den1, den2, den3, minMultiple;

//Taking input fractions for sum
printf("Enter first fraction: ");
scanf("%d%d", &num1, &den1);
printf("Enter second fraction: ");
scanf("%d%d", &num2, &den2);

// Finding LCM of den1 and den2
den3 = lcm(den1,den2);
  
// Changing the fractions to have same denominator
// Numerator of the final fraction obtained
num3 = (num1)*(den3/den1) + (num2)*(den3/den2);
  
//Finding GCD of num3 and den3 to convert result into lowest term
int cd = gcd(num3, den3);
//Converting to lowest term
num3 = num3 / cd;
den3 = den3 / cd;
  
//Printing result
printf("Sum of %d/%d and %d/%d is %d/%d.\n",num1, den1, num2, den2, num3, den3);
return 0;
}

//Finding LCM to compute sum of fractions
int lcm(int n1, int n2){
  
int minMultiple = ( n1 > n2) ? n1 : n2;
//Always TRUE
  
while(1){
  
if (minMultiple % n1 == 0 && minMultiple % n2 == 0){
return minMultiple;
}
  
++minMultiple;
}
}

//Finding greatest common divisor to convert result in lowest term
int gcd(int n1, int n2)
{
if (n2 != 0)
return gcd(n2, n1%n2);
else
return n1;
}

Program Screenshot

1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 4 5 int lcm(int, int); int gcd(int, int); 7. int main() { int numi, num2, num3, den1, den2, den3, minMultiple; //Taking input fractions for sum printf("Enter first fraction: "); scanf("%d%d", &num1, &den1); printf("Enter second fraction: "); scanf("%d%d", &num2, &den2); // Finding LCM of dent and den2 den3 = 1cm(den1,den2); // Changing the fractions to have same denominator // Numerator of the final fraction obtained num3 = (numi) (den/deni) + (num2) (den3 den2); //Finding GCD of num3 and den3 to convert result into Lowest term int cd = gcd(num3, den3); //Converting to Lowest term num3 = num3 / cd; den3 = den3 / cd; //Printing result printf("Sum of %d/%d and %d/%d is %d/%d.\n", num1, den1, num2, den2, num3, den3); return ; 33 } 36 //Finding LCM to compute sum of fractions int lcm(int ni, int n2){ int minMultiple = (n1 > n2) ? n1 : n2; VALOwC TRUIE

den = dens / cd; //Printing result printf("Sum of %d/%d and %d/%d is %d/%d.\n", num1, den1, num2, den2, num3, den3); return 0; 35 //Finding LCM to compute sum of fractions 36 int lcm(int ni, int n2){ int minMultiple = (n1 > n2) ? n1 : n2; //Always TRUE 41 while(1) { 43 if (minMultiple % n1 == 0 && minMultiple % n2 == 0) { return minMultiple; ++minMultiple; 51 //Finding greatest common divisor to convert result in Lowest term 52 int gcd(int ni, int n2) 53 - { if (n2 != 0) return ged(n2, n1%n2); else return n1; se

Program Output

25 i nt cd = gcanums, dens) input Enter first fraction: 5 20 Enter second fraction: 10 80 Sum of 5/20 and 10/80 is 3/8. ... Program finished with exit code o Press ENTER to exit console.

input Enter firsta fraction: 24 Enter second fraction: 36 Sum of 2/4 and 3/6 is 1/1. - ..Program finished with exit code 0 Press ENTER to exit console.