Smallest number divisible by first n numbers

Given a number n find the smallest number evenly divisible by each number 1 to n.

Examples:

Input : n = 4
Output : 12
Explanation : 12 is the smallest numbers divisible
              by all numbers from 1 to 4

Input : n = 10
Output : 2520

Input :  n = 20
Output : 232792560

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)]

Note : In C++ code, the answer quickly exceeds the integer limit, even the long long limit.



Below is the implementation of the logic.

C++

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// C++ program to find smallest number evenly divisible by 
// all numbers 1 to n
#include<bits/stdc++.h>
using namespace std;
  
// Function returns the lcm of first n numbers
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;
}
  
// Driver program to test the above function
int main() 
{
    long long n = 20;
    cout << lcm(n);
    return 0;
}

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Java














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// Java program to find the smallest number evenly divisible by  


// all numbers 1 to n 


  


 class GFG{ 


  


static long gcd(long a, long b) 


{ 


   if(a%b != 0)  


      return gcd(b,a%b); 


   else 


      return b; 


} 


  


// Function returns the lcm of first n numbers 


static long lcm(long n) 


{ 


    long ans = 1;     


    for (long i = 1; i <= n; i++) 


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


    return ans; 


} 


   


// Driver program to test the above function 


public static void main(String []args)  


{ 


    long n = 20; 


    System.out.println(lcm(n)); 


  


} 


} 



















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C#














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// C#  program to find smallest number 


// evenly divisible by  


// all numbers 1 to n  


using System; 


  


public class GFG{ 


    static long gcd(long a, long b)  


{  


if(a%b != 0)  


    return gcd(b,a%b);  


else


    return b;  


}  


  


// Function returns the lcm of first n numbers  


static long lcm(long n)  


{  


    long ans = 1;      


    for (long i = 1; i <= n; i++)  


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


    return ans;  


}  


  


// Driver program to test the above function  


    static public void Main (){ 


        long n = 20;  


        Console.WriteLine(lcm(n));  


    } 


//This code is contributed by akt_mit     


} 



















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Python3














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# Python program to find the smallest number evenly   


# divisible by all number 1 to n  


import math 


    


# Returns the lcm of first n numbers  


def lcm(n):  


    ans = 1


    for i in range(1, n + 1):  


        ans = int((ans * i)/math.gcd(ans, i))          


    return ans  


    


# main  


n = 20


print (lcm(n))  



















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PHP














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<?php 


// Note: This code is not working on GFG-IDE  


// because gmp libraries are not supported 


  


// PHP program to find smallest number  


// evenly divisible by all numbers 1 to n 


  


// Function returns the lcm  


// of first n numbers 


function lcm($n) 


{ 


    $ans = 1;  


    for ($i = 1; $i <= $n; $i++) 


        $ans = ($ans * $i) / (gmp_gcd(strval(ans),  


                                      strval(i))); 


    return $ans; 


} 


  


// Driver Code 


$n = 20; 


echo lcm($n); 


  


// This code is contributed by mits 


?> 



















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Output :


232792560



The above solution works fine for a single input. But if we have multiple inputs, it is a good idea to use Sieve of Eratosthenes to store all prime factors. Please refer below article for Sieve based approach.

LCM of First n Natural Numbers


This article is contributed by Ayush Khanduri. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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