LCM of First n Natural Numbers

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

Examples: 

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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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++

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// 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

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Python3

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# 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

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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 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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