Smallest number divisible by first n numbers - GeeksforGeeks

Smallest number divisible by first n numbers - GeeksforGeeks
Given a number n find the smallest number evenly divisible by each number 1 to n.

If you observe carefully the ans must be the LCM of the numbers 1 to n.
To find LCM of numbers from 1 to n –

1. Initialize ans = 1.
2. Iterate over all the numbers from i = 1 to i = n.
   At the i’th iteration ans = LCM(1, 2, …….., i). This can be done easily as LCM(1, 2, …., i) = LCM(ans, i).
   Thus at i’th iteration we just have to do –

   ans = LCM(ans, i) 
            = ans * i / gcd(ans, i) [Using the below property,
                                    a*b = gcd(a,b) * lcm(a,b)]

long long lcm(long long n)

{

    long long ans = 1;   

    for (long long i = 1; i <= n; i++)

        ans = (ans * i)/(__gcd(ans, i));

    return ans;

}

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