C Program to Find LCM of n Numbers

In this tutorial, you will learn and get code about finding the LCM of n numbers. Or you can says that, here the program is created to find LCM of an array. But before going through the program, I recommend you to visit Find LCM of Two Numbers once (if not visited). Because, the program given here is relatable to the program in that article.

Find LCM of n Numbers in C

The question is write a program in C to find out LCM of n numbers (a given array). The answer to this question is:

#include<stdio.h>
#include<conio.h>
int main()
{
    int arr[10], n, mp, i, count;
    printf("Enter the Size: ");
    scanf("%d", &n);
    printf("Enter %d Numbers: ", n);
    for(i=0; i<n; i++)
        scanf("%d", &arr[i]);
    i=0;
    mp = arr[i];
    while(i<n)
    {
        if(mp<arr[i])
            mp = arr[i];
        i++;
    }
    while(1)
    {
        i=0;
        count=0;
        while(i<n)
        {
            if(mp%arr[i]==0)
                count++;
            i++;
        }
        if(count==n)
            break;
        else
            mp++;
    }
    printf("\nLCM(");
    for(i=0; i<(n-1); i++)
        printf("%d,", arr[i]);
    printf("%d) = %d", arr[i], mp);
    getch();
    return 0;
}

This program was build and run under Code::Blocks IDE. Here is its sample run:

Now supply 5 as size and enter 5 numbers say 10, 12, 15, 20, 30. Then press ENTER key to see the following output:

Program Explained

The statement:

while(i<n)
{
    if(mp%arr[i]==0)
        count++;
    i++;
}

is used to check whether value at mp is divisible by all the numbers or not. If it is divisible by any number, then increment the value of count. After exiting this loop, checks whether count is equal to the array size or not. If it is equal, then mp is divisible by all the numbers. Because every time the condition of if blocks evaluates to be true and program flow goes inside the block and increments the value of count.

For example, let's suppose the size of array is 5 and the elements are 10, 12, 15, 20, 30. Then program flow goes like:

• Element at index 0 (first number) gets initialized to mp
• Compares it with next index's element (second number)
• If value of mp is less than the number at next index, then initialize its value to mp
• This block of code is created only to assign the largest element of the array to mp, to start from highest number
• Using 1 as while loop's condition, always evaluates to be true
• Therefore, this loop continue running until break keyword occurs
• Inside the loop, 0 is initialized to count and i
• Create a while loop to run 5 number of times
• Inside the loop, check whether value at mp is divisible by number at current index or not
• If it is, then increment the value of count and goes for next element
• After checking for all the 5 elements
• Check whether count is equal to the size of array or not
• If it is equal, then it means that the condition mp%arr[i]==0 evaluates to be true, 5 times. And it means, mp is divisible by all the 5 numbers
• Therefore use the break statement and exit from the loop
• And print its value as LCM of given 5 numbers
• And if it is not equal, then increment the value of mp and continue the same process
• Do until it is divisible by all the 5 numbers. That's it

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