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++
// 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 numbersint maxLCM(int n){ return (n * (n - 1));}// Driver codeint main(){ int n = 3; cout << maxLCM(n); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to return the maximum LCM// among all the pairs(i, j) of// first n natural numbersstatic int maxLCM(int n){ return (n * (n - 1));}// Driver codepublic 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 numbersstatic int maxLCM(int n){ return (n * (n - 1));}// Driver codepublic 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 numbersfunction maxLCM(n){ return (n * (n - 1));}// Driver codevar n = 3;document.write(maxLCM(n));</script> |
Output:
6
Time Complexity: O(1)
Auxiliary Space: O(1)
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