Given a positive number N. The task is to find the maximum sum of distinct numbers such that the LCM of all these numbers is equal to N.
Examples:
Input : 2 Output : 3 The distinct numbers you can have are just 1 and 2 and their sum is equal to 3. Input : 5 Output : 6
As the LCM of all the numbers is N. So all the numbers must be the divisors of N and all the numbers are distinct so answer must be the sum of all the divisors of N. To find all the divisors efficiently refer to article https://www.geeksforgeeks.org/find-all-divisors-of-a-natural-number-set-2/
Here are the steps to solve this problem :
- Find all the divisors of ‘n’ by iterating from 1 to the square root of the n.
- If i!= n/i, add n/i to max sum as well as each divisor i to max_sum.
- Return max_sum as the final result.
Below is the implementation of the above approach.
C++
// C++ program to find the max sum of // numbers whose lcm is n #include<bits/stdc++.h> using namespace std; // Returns maximum sum of numbers with // LCM as N int maxSumLCM( int n) { int max_sum = 0; // Initialize result // Finding a divisor of n and adding // it to max_sum for ( int i=1; i*i<=n; i++) { if (n%i == 0) { max_sum += i; if (n/i != i) max_sum += (n/i); } } return max_sum; } // Driver code int main() { int n = 2; cout << maxSumLCM(n) << endl; return 0; } |
Java
// Java program to find the max sum of // numbers whose lcm is n class MaxSum { // Returns maximum sum of numbers with // LCM as N static int maxSumLCM( int n) { int max_sum = 0 ; // Initialize result // Finding a divisor of n and adding // it to max_sum for ( int i= 1 ; i*i<=n; i++) { if (n%i == 0 ) { max_sum += i; if (n/i != i) max_sum += (n/i); } } return max_sum; } // main function public static void main (String[] args) { int n = 2 ; System.out.println(maxSumLCM(n)); } } |
Python3
# Python3 program to find the max sum of # numbers whose lcm is n # Returns maximum sum of numbers with # LCM as N def maxSumLCM(n) : # Initialize result max_sum = 0 # Finding a divisor of n and adding # it to max_sum i = 1 while (i * i< = n ): if (n % i = = 0 ) : max_sum = max_sum + i if (n / / i ! = i) : max_sum = max_sum + (n / / i) i = i + 1 return max_sum # Driver code n = 2 print (maxSumLCM(n)) # This code is contributed by Nikita Tiwari. |
C#
// C# program to find the max sum // of numbers whose lcm is n using System; class MaxSum { // Returns maximum sum of // numbers with LCM as N static int maxSumLCM( int n) { // Initialize result int max_sum = 0; // Finding a divisor of n and // adding it to max_sum for ( int i = 1; i * i <= n; i++) { if (n % i == 0) { max_sum += i; if (n / i != i) max_sum += (n / i); } } return max_sum; } // Driver Code public static void Main (String[] args) { int n = 2; Console.Write(maxSumLCM(n)); } } // This code is contributed by parashar.. |
PHP
<?php // PHP program to find // the max sum of numbers // whose lcm is n // Returns maximum sum // of numbers with // LCM as N function maxSumLCM( $n ) { // Initialize result $max_sum = 0; // Finding a divisor // of n and adding // it to max_sum for ( $i = 1; $i * $i <= $n ; $i ++) { if ( $n % $i == 0) { $max_sum += $i ; if ( $n / $i != $i ) $max_sum += ( $n / $i ); } } return $max_sum ; } // Driver code $n = 2; echo MaxSumLCM( $n ); // This code is contributed // by ajit ?> |
Javascript
<script> // Javascript program to find the max sum of // numbers whose lcm is n // Returns maximum sum of numbers with // LCM as N function maxSumLCM(n) { let max_sum = 0; // Initialize result // Finding a divisor of n and adding // it to max_sum for (let i=1; i*i<=n; i++) { if (n%i == 0) { max_sum += i; if (n/i != i) max_sum += (n/i); } } return max_sum; } // Driver code let n = 2; document.write(maxSumLCM(n) + "<br>" ); // This code is contributed by Mayank Tyagi </script> |
3
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
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