LCM of First n Natural Numbers (original) (raw)

Last Updated : 3 Jun, 2026

Given a number **n, find an integer denoting the smallest number divisible by each number from **1 to **n.

**Examples:

**Input: n = 3
**Output: 6
**Explanation: 6 is the smallest number divisible by 1, 2 and 3.

**Input: n = 6
**Output: 60
**Explanation: 60 is the smallest number divisible by all from 1 to 6.

Try It Yourself

Table of Content

• Using LCM of Every Number - O(n log n) Time O(log n) Space
• Using Prime Powers / Sieve - O(n log log n) Time O(n) Space

Using LCM of Every Number - O(n log n) Time O(log n) Space

The idea is to start with res = 1 and iteratively compute the LCM of res and every number from 1 to n. We use: LCM(a, b) = (a * b) / GCD(a, b), After processing all numbers, res becomes the LCM of all numbers from 1 to n.

C++ `

#include #include using namespace std;

// Function to find GCD long long gcd(long long a, long long b) { if (b == 0) return a;

return gcd(b, a % b);

}

long long getSmallestDivNum(long long n) {

long long res = 1;

// Find LCM of all numbers from 1 to n
for (long long i = 1; i <= n; i++)
{
    res = (res * i) / gcd(res, i);
}

return res;

}

// Driver code int main() { long long n = 6;

cout << getSmallestDivNum(n);

return 0;

}

Java

import java.util.*;

public class GfG { // Function to find GCD public static long gcd(long a, long b) { if (b == 0) return a;

    return gcd(b, a % b);
}

public static long getSmallestDivNum(long n)
{

    long res = 1;

    // Find LCM of all numbers from 1 to n
    for (long i = 1; i <= n; i++) {
        res = (res * i) / gcd(res, i);
    }

    return res;
}

// Driver code
public static void main(String[] args)
{
    long n = 6;

    System.out.println(getSmallestDivNum(n));
}

}

Python

def gcd(a, b): if b == 0: return a

return gcd(b, a % b)

def getSmallestDivNum(n):

res = 1

# Find LCM of all numbers from 1 to n
for i in range(1, n + 1):
    res = (res * i) // gcd(res, i)

return res

Driver code

if name == 'main': n = 6

print(getSmallestDivNum(n))

C#

using System;

public class GfG { // Function to find GCD public static long gcd(long a, long b) { if (b == 0) return a;

    return gcd(b, a % b);
}

public static long getSmallestDivNum(long n)
{

    long res = 1;

    // Find LCM of all numbers from 1 to n
    for (long i = 1; i <= n; i++) {
        res = (res * i) / gcd(res, i);
    }

    return res;
}

// Driver code
public static void Main()
{
    long n = 6;

    Console.WriteLine(getSmallestDivNum(n));
}

}

JavaScript

function gcd(a, b) { if (b === 0) return a;

return gcd(b, a % b);

}

function getSmallestDivNum(n) {

let res = 1;

// Find LCM of all numbers from 1 to n
for (let i = 1; i <= n; i++) {
    res = (res * i) / gcd(res, i);
}

return res;

}

// Driver code let n = 6;

console.log(getSmallestDivNum(n));

`

**Time Complexity: O(n log n)
**Auxiliary Space: O(log n)

Using Prime Powers / Sieve - O(n log log n) Time O(n) Space

The idea is to use the fact that the LCM of numbers from 1 to n contains the highest power of every prime that does not exceed n. For every prime p ≤ n:

• Find the largest power p^k such that p^k ≤ n.
• Multiply all such powers together.

**Let us understand with example:
For n = 6:

• Highest power of 2 ≤ 6 is 2^2 which is 4
• Highest power of 3 ≤ 6 is 3*1 which 3
• Highest power of 5 ≤ 6 is 5*1 which is 5

**LCM = 4 × 3 × 5 = 60

C++ `

#include #include using namespace std;

long long getSmallestDivNum(long long n) {

vector<bool> isPrime(n + 1, true);

// Build Sieve of Eratosthenes
isPrime[0] = false;

if (n >= 1)
    isPrime[1] = false;

for (long long i = 2; i * i <= n; i++)
{
    if (isPrime[i])
    {
        for (long long j = i * i; j <= n; j += i)
        {
            isPrime[j] = false;
        }
    }
}

long long lcm = 1;

// Process every prime
for (long long i = 2; i <= n; i++)
{
    if (isPrime[i])
    {
        long long power = i;

        // Find highest power of prime <= n
        while (power <= n / i)
        {
            power *= i;
        }

        lcm *= power;
    }
}

return lcm;

}

// Driver code int main() { long long n = 6;

cout << getSmallestDivNum(n);

return 0;

}

Java

import java.util.Arrays;

public class GfG { public static long getSmallestDivNum(long n) { boolean[] isPrime = new boolean[(int)(n + 1)]; Arrays.fill(isPrime, true);

    // Build Sieve of Eratosthenes
    isPrime[0] = false;

    if (n >= 1)
        isPrime[1] = false;

    for (long i = 2; i * i <= n; i++) {
        if (isPrime[(int)i]) {
            for (long j = i * i; j <= n; j += i) {
                isPrime[(int)j] = false;
            }
        }
    }

    long lcm = 1;

    // Process every prime
    for (long i = 2; i <= n; i++) {
        if (isPrime[(int)i]) {
            long power = i;

            // Find highest power of prime <= n
            while (power <= n / i) {
                power *= i;
            }

            lcm *= power;
        }
    }

    return lcm;
}

public static void main(String[] args)
{
    long n = 6;
    System.out.println(getSmallestDivNum(n));
}

}

Python

def getSmallestDivNum(n): isPrime = [True] * (n + 1)

# Build Sieve of Eratosthenes
isPrime[0] = False

if n >= 1:
    isPrime[1] = False

for i in range(2, int(n**0.5) + 1):
    if isPrime[i]:
        for j in range(i * i, n + 1, i):
            isPrime[j] = False

lcm = 1

# Process every prime
for i in range(2, n + 1):
    if isPrime[i]:
        power = i

        # Find highest power of prime <= n
        while power <= n // i:
            power *= i

        lcm *= power

return lcm

Driver code

if name == "main": n = 6

print(getSmallestDivNum(n))

C#

using System;

public class GfG { public static long getSmallestDivNum(long n) { bool[] isPrime = new bool[n + 1]; for (int i = 0; i <= n; i++) isPrime[i] = true;

    // Build Sieve of Eratosthenes
    isPrime[0] = false;

    if (n >= 1)
        isPrime[1] = false;

    for (long i = 2; i * i <= n; i++) {
        if (isPrime[i]) {
            for (long j = i * i; j <= n; j += i) {
                isPrime[j] = false;
            }
        }
    }

    long lcm = 1;

    // Process every prime
    for (long i = 2; i <= n; i++) {
        if (isPrime[i]) {
            long power = i;

            // Find highest power of prime <= n
            while (power <= n / i) {
                power *= i;
            }

            lcm *= power;
        }
    }

    return lcm;
}

public static void Main()
{
    long n = 6;
    Console.WriteLine(getSmallestDivNum(n));
}

}

JavaScript

function getSmallestDivNum(n) { let isPrime = Array(n + 1).fill(true);

// Build Sieve of Eratosthenes
isPrime[0] = false;

if (n >= 1)
    isPrime[1] = false;

for (let i = 2; i * i <= n; i++) {
    if (isPrime[i]) {
        for (let j = i * i; j <= n; j += i) {
            isPrime[j] = false;
        }
    }
}

let lcm = 1;

// Process every prime
for (let i = 2; i <= n; i++) {
    if (isPrime[i]) {
        let power = i;

        // Find highest power of prime <= n
        while (power <= Math.floor(n / i)) {
            power *= i;
        }

        lcm *= power;
    }
}

return lcm;

}

// Driver code let n = 6; console.log(getSmallestDivNum(n));

`

**Time Complexity: O(n log log n)
**Auxiliary Space: O(n)