LCM of First n Natural Numbers - GeeksforGeeks

27 Mar, 2022
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 have discussed a simple solution in below article. Smallest number divisible by first n numbersThe 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’.  
Initialize lcm variable with 1Generate a Sieve of Eratosthenes (bool vector isPrime) of length 10^6 (ideally must be equal to no. of digits in factorial)Now, for each number in bool vector isPrime, if the number is prime (isPrime[i] is true), find the maximum number which is less than the given number and equal to power of the prime.Then multiply this number with lcm variable.Repeat step 3 and 4 until prime is less than the given number.
Initialize lcm variable with 1
Generate a Sieve of Eratosthenes (bool vector isPrime) of length 10^6 (ideally must be equal to no. of digits in factorial)
Now, for each number in bool vector isPrime, if the number is prime (isPrime[i] is true), find the maximum number which is less than the given number and equal to power of the prime.
Then multiply this number with lcm variable.
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++
Java
Python3
C#
Javascript
PHP
// C++ program to find LCM of First N Natural Numbers.#include <bits/stdc++.h>#define MAX 100000using namespace std; vector<bool> isPrime (MAX, true);// utility function for sieve of sieve of Eratosthenes void sieve(){         for (int i = 2; i * i <= MAX; i++)    {        if (isPrime[i] == true)            for (int j = i*i; j<= MAX; j+=i)                isPrime[j] = false;    }} // Function to find LCM of first n Natural Numberslong long LCM(int n){    long long lcm = 1;    int i=2;       while(i<=n)    {        if(isPrime[i]){            int pp = i;            while (pp * i <= n)                pp = pp * i;            lcm *= pp;        }        i++;    }    return lcm;} // Driver codeint main(){    // build sieve    sieve();    int N = 7;    // Function call    cout << LCM(N);    return 0;}
// 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 program to find LCM of# First N Natural Numbers.MAX = 100000 # array to store all prime less# than and equal to 10^6primes = [] # 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 codesieve()N = 7 # Function callprint(LCM(N)) # This code is contributed by mits
// 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
<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 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 Eratosthenesfunction 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, $i);} // Function to find LCM of// first n Natural Numbersfunction LCM($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 $lcm;} // Driver codesieve();$N = 7; // Function callecho LCM($N); // This code is contributed by mits?>
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++
Java
Python3
C#
Javascript
// C++ program to find LCM of First N Natural Numbers.#include <bits/stdc++.h>using namespace std; // to calculate hcfint 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 codeint 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 program to find LCM of First N Natural Numberspublic 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 program to find LCM# of First N Natural Numbers. # To calculate hcfdef 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 coden = 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# 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.
<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>
420
Time complexity : O(nlog n)
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