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C Exercises: Determine the LCM of two numbers using HCF

C For Loop: Exercise-44 with Solution

Write a program in C to find LCM of any two numbers using HCF.

Pictorial Presentation:

Determine the LCM of two numbers using HCF

Sample Solution:

C Code:

#include <stdio.h>

void main()  
{  
    int i, n1, n2, j, hcf=1,lcm;  


     printf("\n\n  LCM of two numbers:\n ");
     printf("----------------------\n");


    printf("Input 1st number for LCM: ");  
    scanf("%d", &n1);  
    printf("Input 2nd number for LCM: ");  
    scanf("%d", &n2); 
  
    j = (n1<n2) ? n1 : n2;  
  
    for(i=1; i<=j; i++)  
    {  

        if(n1%i==0 && n2%i==0)  
        {  
            hcf = i;  
        }  
    }  
/* We know  the multiplication of HCF and LCM is equivalant
to the multiplication of these two numbers.*/
    lcm=(n1*n2)/hcf;
  
    printf("\nThe LCM of %d and %d is : %d\n\n", n1, n2, lcm);  

} 

Sample Output:

  LCM of two numbers:                                                                                         
 ----------------------                                                                                       
Input 1st number for LCM: 15                                                                                  
Input 2nd number for LCM: 20                                                                                  
                                                                                                              
The LCM of 15 and 20 is : 60

Flowchart:

Flowchart : Determine the LCM of two numbers using HCF.

C Programming Code Editor:

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Previous: Write a C program to find HCF (Highest Common Factor) of two numbers.
Next: Write a program in C to find LCM of any two numbers.

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C Programming: Tips of the Day

Printing hexadecimal characters in C:

You are seeing the ffffff because char is signed on your system. In C, vararg functions such as printf will promote all integers smaller than int to int. Since char is an integer (8-bit signed integer in your case), your chars are being promoted to int via sign-extension.

Since c0 and 80 have a leading 1-bit (and are negative as an 8-bit integer), they are being sign-extended while the others in your sample don't.

char    int
c0 -> ffffffc0
80 -> ffffff80
61 -> 00000061
Here's a solution:
char ch = 0xC0;
printf("%x", ch & 0xff);

This will mask out the upper bits and keep only the lower 8 bits that you want.

Ref : https://bit.ly/3vOLizM