Skip to content
Related Articles
Open in App
Not now

Related Articles

Smallest number divisible by first n numbers

Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 30 Dec, 2022
Improve Article
Save Article

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)]
  1.  

Note : In C++ code, the answer quickly exceeds the integer limit, even the long long limit.
Below is the implementation of the logic. 
 

C++




// 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;
}

Java




// 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));
 
}
}

C#




// 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   
}

Python3




# 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))

PHP




<?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
?>

Javascript




<script>
 
// Javascript program to find the smallest number evenly divisible by
// all numbers 1 to n
 
function gcd(a, b)
{
   if(a%b != 0)
      return gcd(b,a%b);
   else
      return b;
}
   
// Function returns the lcm of first n numbers
function lcm(n)
{
    let ans = 1;   
    for (let i = 1; i <= n; i++)
        ans = (ans * i)/(gcd(ans, i));
    return ans;
}
       
// function call
     
    let n = 20;
    document.write(lcm(n));
     
</script>

Output

232792560

Time Complexity: O(n log2n) as the complexity of _gcd(a,b) in c++ is log2n  and that is running n times in a loop.
Auxiliary Space: O(1)
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
 


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!