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Maximum LCM among all pairs (i, j) of first N natural numbers

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  • Last Updated : 10 Mar, 2022
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Given a positive integer N > 1, the task is to find the maximum LCM among all the pairs (i, j) such that i < j ≤ N.
Examples: 
 

Input: N = 3 
Output:
LCM(1, 2) = 2 
LCM(1, 3) = 3 
LCM(2, 3) = 6
Input: N = 4 
Output: 12 
 

 

Approach: Since the LCM of two consecutive elements is equal to their multiples then it is obvious that the maximum LCM will be of the pair (N, N – 1) i.e. (N * (N – 1)).
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
int maxLCM(int n)
{
    return (n * (n - 1));
}
 
// Driver code
int main()
{
    int n = 3;
 
    cout << maxLCM(n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
static int maxLCM(int n)
{
    return (n * (n - 1));
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
 
    System.out.println(maxLCM(n));
}
}
 
// This code is contributed by Code_Mech

Python3




# Python3 implementation of the approach
 
# Function to return the maximum LCM
# among all the pairs(i, j) of
# first n natural numbers
def maxLCM(n) :
 
    return (n * (n - 1));
 
# Driver code
if __name__ == "__main__" :
 
    n = 3;
 
    print(maxLCM(n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
     
class GFG
{
     
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
static int maxLCM(int n)
{
    return (n * (n - 1));
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3;
 
    Console.WriteLine(maxLCM(n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
function maxLCM(n)
{
    return (n * (n - 1));
}
 
// Driver code
var n = 3;
document.write(maxLCM(n));
 
</script>

Output: 

6

 

Time Complexity: O(1)

Auxiliary Space: O(1)
 


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