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

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: 6
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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

6

Time Complexity: O(1)



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