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LCM of First n Natural Numbers

  • Difficulty Level : Medium
  • Last Updated : 29 Sep, 2021

Given a number n such that 1 <= N <= 10^6, the Task is to Find the LCM of First n Natural Numbers. 

Examples: 

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Input : n = 5
Output : 60

Input : n = 6
Output : 60

Input : n = 7
Output : 420 

We strongly recommend that you click here and practice it, before moving on to the solution.

We have discussed a simple solution in below article. 
Smallest number divisible by first n numbers
The above solution works fine for single input. But if we have multiple inputs, it is a good idea to use Sieve of Eratosthenes to store all prime factors. As we know if LCM(a, b) = X so any prime factor of a or b will also be the prime factor of ‘X’.  

  1. Initialize lcm variable with 1
  2. Generate all prime number less then 10^6 and store in Array prime by using Sieve of Eratosthenes.
  3. Find the maximum number which is less than the given number and equal to power of the prime.
  4. Then multiply this number with lcm variable.
  5. Repeat step 3 and 4 until prime is less than the given number.

Illustration: 



For example, if n = 10 
8 will be the first number which is equal to 2^3
then 9 which is equal to 3^2
then 5 which is equal to 5^1
then 7 which is equal to 7^1
Finally, we multiply those numbers 8*9*5*7 = 2520

Below is the implementation of the above idea.  

C++




// C++ program to find LCM of First N Natural Numbers.
#include <bits/stdc++.h>
#define MAX 100000
using namespace std;
 
// array to store all prime less than and equal to 10^6
vector<int> primes;
 
// utility function for sieve of sieve of Eratosthenes
void sieve()
{
    bool isComposite[MAX] = { false };
    for (int i = 2; i * i <= MAX; i++)
    {
        if (isComposite[i] == false)
            for (int j = 2; j * i <= MAX; j++)
                isComposite[i * j] = true;
    }
 
    // Store all prime numbers in vector primes[]
    for (int i = 2; i <= MAX; i++)
        if (isComposite[i] == false)
            primes.push_back(i);
}
 
// Function to find LCM of first n Natural Numbers
long long LCM(int n)
{
    long long lcm = 1;
    for (int i = 0;
         i < primes.size() && primes[i] <= n;
         i++)
    {
        // Find the highest power of prime, primes[i]
        // that is less than or equal to n
        int pp = primes[i];
        while (pp * primes[i] <= n)
            pp = pp * primes[i];
 
        // multiply lcm with highest power of prime[i]
        lcm *= pp;
        lcm %= 1000000007;
    }
    return lcm;
}
 
// Driver code
int main()
{
    // build sieve
    sieve();
    int N = 7;
   
    // Function call
    cout << LCM(N);
    return 0;
}

Java




// Java program to find LCM of First N Natural Numbers.
import java.util.*;
 
class GFG
{
    static int MAX = 100000;
 
    // array to store all prime less than and equal to 10^6
    static ArrayList<Integer> primes
        = new ArrayList<Integer>();
    // utility function for sieve of sieve of Eratosthenes
    static void sieve()
    {
        boolean[] isComposite = new boolean[MAX + 1];
        for (int i = 2; i * i <= MAX; i++)
        {
            if (isComposite[i] == false)
                for (int j = 2; j * i <= MAX; j++)
                    isComposite[i * j] = true;
        }
 
        // Store all prime numbers in vector primes[]
        for (int i = 2; i <= MAX; i++)
            if (isComposite[i] == false)
                primes.add(i);
    }
 
    // Function to find LCM of first n Natural Numbers
    static long LCM(int n)
    {
        long lcm = 1;
        for (int i = 0;
             i < primes.size() && primes.get(i) <= n;
             i++)
        {
            // Find the highest power of prime, primes[i]
            // that is less than or equal to n
            int pp = primes.get(i);
            while (pp * primes.get(i) <= n)
                pp = pp * primes.get(i);
 
            // multiply lcm with highest power of prime[i]
            lcm *= pp;
            lcm %= 1000000007;
        }
        return lcm;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        sieve();
        int N = 7;
       
        // Function call
        System.out.println(LCM(N));
    }
}
// This code is contributed by mits

Python3




# Python3 program to find LCM of
# First N Natural Numbers.
MAX = 100000
 
# array to store all prime less
# than and equal to 10^6
primes = []
 
# utility function for
# sieve of Eratosthenes
 
 
def sieve():
 
    isComposite = [False]*(MAX+1)
    i = 2
    while (i * i <= MAX):
        if (isComposite[i] == False):
            j = 2
            while (j * i <= MAX):
                isComposite[i * j] = True
                j += 1
        i += 1
 
    # Store all prime numbers in
    # vector primes[]
    for i in range(2, MAX+1):
        if (isComposite[i] == False):
            primes.append(i)
 
# Function to find LCM of
# first n Natural Numbers
 
 
def LCM(n):
 
    lcm = 1
    i = 0
    while (i < len(primes) and primes[i] <= n):
        # Find the highest power of prime,
        # primes[i] that is less than or
        # equal to n
        pp = primes[i]
        while (pp * primes[i] <= n):
            pp = pp * primes[i]
 
        # multiply lcm with highest
        # power of prime[i]
        lcm *= pp
        lcm %= 1000000007
        i += 1
    return lcm
 
 
# Driver code
sieve()
N = 7
 
# Function call
print(LCM(N))
 
# This code is contributed by mits

C#




// C# program to find LCM of First N
// Natural Numbers.
using System.Collections;
using System;
 
class GFG {
    static int MAX = 100000;
 
    // array to store all prime less than
    // and equal to 10^6
    static ArrayList primes = new ArrayList();
 
    // utility function for sieve of
    // sieve of Eratosthenes
    static void sieve()
    {
        bool[] isComposite = new bool[MAX + 1];
        for (int i = 2; i * i <= MAX; i++)
        {
            if (isComposite[i] == false)
                for (int j = 2; j * i <= MAX; j++)
                    isComposite[i * j] = true;
        }
 
        // Store all prime numbers in vector primes[]
        for (int i = 2; i <= MAX; i++)
            if (isComposite[i] == false)
                primes.Add(i);
    }
 
    // Function to find LCM of first
    // n Natural Numbers
    static long LCM(int n)
    {
        long lcm = 1;
        for (int i = 0;
             i < primes.Count && (int)primes[i] <= n;
             i++)
        {
            // Find the highest power of prime, primes[i]
            // that is less than or equal to n
            int pp = (int)primes[i];
            while (pp * (int)primes[i] <= n)
                pp = pp * (int)primes[i];
 
            // multiply lcm with highest power of prime[i]
            lcm *= pp;
            lcm %= 1000000007;
        }
        return lcm;
    }
 
    // Driver code
    public static void Main()
    {
        sieve();
        int N = 7;
       
        // Function call
        Console.WriteLine(LCM(N));
    }
}
 
// This code is contributed by mits

Javascript




<script>
    // Javascript program to find LCM of First N
    // Natural Numbers.
     
    let MAX = 100000;
  
    // array to store all prime less than
    // and equal to 10^6
    let primes = [];
  
    // utility function for sieve of
    // sieve of Eratosthenes
    function sieve()
    {
        let isComposite = new Array(MAX + 1);
        isComposite.fill(false);
        for (let i = 2; i * i <= MAX; i++)
        {
            if (isComposite[i] == false)
                for (let j = 2; j * i <= MAX; j++)
                    isComposite[i * j] = true;
        }
  
        // Store all prime numbers in vector primes[]
        for (let i = 2; i <= MAX; i++)
            if (isComposite[i] == false)
                primes.push(i);
    }
  
    // Function to find LCM of first
    // n Natural Numbers
    function LCM(n)
    {
        let lcm = 1;
        for (let i = 0;
             i < primes.length && primes[i] <= n;
             i++)
        {
            // Find the highest power of prime, primes[i]
            // that is less than or equal to n
            let pp = primes[i];
            while (pp * primes[i] <= n)
                pp = pp * primes[i];
  
            // multiply lcm with highest power of prime[i]
            lcm *= pp;
            lcm %= 1000000007;
        }
        return lcm;
    }
     
    sieve();
    let N = 7;
 
    // Function call
    document.write(LCM(N));
 
// This code is contributed by decode2207.
</script>

PHP




<?php
// PHP program to find LCM of
// First N Natural Numbers.
$MAX = 100000;
 
// array to store all prime less
// than and equal to 10^6
$primes = array();
 
// utility function for
// sieve of Eratosthenes
function sieve()
{
    global $MAX, $primes;
    $isComposite = array_fill(0, $MAX, false);
    for ($i = 2; $i * $i <= $MAX; $i++)
    {
        if ($isComposite[$i] == false)
            for ($j = 2; $j * $i <= $MAX; $j++)
                $isComposite[$i * $j] = true;
    }
 
    // Store all prime numbers in
    // vector primes[]
    for ($i = 2; $i <= $MAX; $i++)
        if ($isComposite[$i] == false)
            array_push($primes, $i);
}
 
// Function to find LCM of
// first n Natural Numbers
function LCM($n)
{
    global $MAX, $primes;
    $lcm = 1;
    for ($i = 0; $i < count($primes) &&
                 $primes[$i] <= $n; $i++)
    {
        // Find the highest power of prime,
        // primes[i] that is less than or
        // equal to n
        $pp = $primes[$i];
        while ($pp * $primes[$i] <= $n)
            $pp = $pp * $primes[$i];
 
        // multiply lcm with highest
        // power of prime[i]
        $lcm *= $pp;
        $lcm %= 1000000007;
    }
    return $lcm;
}
 
// Driver code
sieve();
$N = 7;
 
// Function call
echo LCM($N);
 
// This code is contributed by mits
?>
Output
420

Another Approach:

 The idea is that if the number is less than 3 then return number. If the number is greater than 2 then find LCM of n,n-1

  • Lets say x=LCM(n,n-1)
  • again x=LCM(x,n-2)
  • again x=LCM(x,n-3) …
  • .
  • .
  • again x=LCM(x,1) …

now the result is x.

For finding LCM(a,b) we use a function hcf(a,b) whichwill return HCF of (a,b)

We know that LCM(a,b)= (a*b)/HCF(a,b)

Illustration: 

For example, if n = 7 
function call lcm(7,6)
now lets say a=7 , b=6

Now , b!= 1 Hence 
a=lcm(7,6) = 42 and b=6-1=5

function call lcm(42,5)
a=lcm(42,5) = 210 and b=5-1=4

function call lcm(210,4)
a=lcm(210,4) = 420 and b=4-1=3

function call lcm(420,3)
a=lcm(420,3) = 420 and b=3-1=2

function call lcm(420,2)
a=lcm(420,2) = 420 and b=2-1=1

Now b=1
Hence return a=420

Below is the implementation of the above approach

C++




// C++ program to find LCM of First N Natural Numbers.
#include <bits/stdc++.h>
using namespace std;
 
// to calculate hcf
int hcf(int a, int b)
{
    if (b == 0)
        return a;
    return hcf(b, a % b);
}
 
 
int findlcm(int a,int b)
{
    if (b == 1)
       
        // lcm(a,b)=(a*b)/hcf(a,b)
        return a;
   
    // assign a=lcm of n,n-1
    a = (a * b) / hcf(a, b);
   
    // b=b-1
    b -= 1;
    return findlcm(a, b);
}
 
// Driver code
int main()
{
    int n = 7;
    if (n < 3)
        cout << n; // base case
    else
        
        // Function call
        // pass n,n-1 in function to find LCM of first n natural
        // number
        cout << findlcm(n, n - 1);
     
    return 0;
}
 
// contributed by ajaykr00kj

Java




// Java program to find LCM of First N Natural Numbers
public class Main
{
  // to calculate hcf
  static int hcf(int a, int b)
  {
    if (b == 0)
      return a;
    return hcf(b, a % b);
  }
 
 
  static int findlcm(int a,int b)
  {
    if (b == 1)
 
      // lcm(a,b)=(a*b)/hcf(a,b)
      return a;
 
    // assign a=lcm of n,n-1
    a = (a * b) / hcf(a, b);
 
    // b=b-1
    b -= 1;
    return findlcm(a, b);
  }
 
  // Driver code.
  public static void main(String[] args)
  {
    int n = 7;
    if (n < 3)
      System.out.print(n); // base case
    else
 
      // Function call
      // pass n,n-1 in function to find LCM of first n natural
      // number
      System.out.print(findlcm(n, n - 1));
  }
}
 
// This code is contributed by divyeshrabadiya07.

Python3




# Python3 program to find LCM
# of First N Natural Numbers.
 
# To calculate hcf
def hcf(a, b):
     
    if (b == 0):
        return a
         
    return hcf(b, a % b)
     
def findlcm(a, b):
     
    if (b == 1):
         
        # lcm(a,b)=(a*b)//hcf(a,b)
        return a
     
    # Assign a=lcm of n,n-1
    a = (a * b) // hcf(a, b)
     
    # b=b-1
    b -= 1
     
    return findlcm(a, b)
 
# Driver code
n = 7
 
if (n < 3):
    print(n)
else:
     
    # Function call
    # pass n,n-1 in function
    # to find LCM of first n
    # natural number
    print(findlcm(n, n - 1))
 
# This code is contributed by Shubham_Singh

C#




// C# program to find LCM of First N Natural Numbers.
using System;
class GFG {
 
  // to calculate hcf
  static int hcf(int a, int b)
  {
    if (b == 0)
      return a;
    return hcf(b, a % b);
  }
 
  static int findlcm(int a,int b)
  {
    if (b == 1)
 
      // lcm(a,b)=(a*b)/hcf(a,b)
      return a;
 
    // assign a=lcm of n,n-1
    a = (a * b) / hcf(a, b);
 
    // b=b-1
    b -= 1;
    return findlcm(a, b);
  }
 
  // Driver code
  static void Main() {
    int n = 7;
    if (n < 3)
      Console.Write(n); // base case
    else
 
      // Function call
      // pass n,n-1 in function to find LCM of first n natural
      // number
      Console.Write(findlcm(n, n - 1));
  }
}
 
// This code is contributed by divyesh072019.

Javascript




<script>
 
    // Javascript program to find LCM of First N Natural Numbers.
     
    // to calculate hcf
    function hcf(a, b)
    {
        if (b == 0)
            return a;
        return hcf(b, a % b);
    }
 
 
    function findlcm(a,b)
    {
        if (b == 1)
 
            // lcm(a,b)=(a*b)/hcf(a,b)
            return a;
 
        // assign a=lcm of n,n-1
        a = (a * b) / hcf(a, b);
 
        // b=b-1
        b -= 1;
        return findlcm(a, b);
    }
     
    let n = 7;
    if (n < 3)
        document.write(n); // base case
    else
         
        // Function call
        // pass n,n-1 in function to find LCM of first n natural
        // number
        document.write(findlcm(n, n - 1));
     
</script>
Output
420

Time complexity : O(nlog n)

This article is contributed by Kuldeep Singh (kulli_d_coder). 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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