# Find the maximum sum (a+b) for a given input integer N satisfying the given condition

Given an integer N, the task is to find the largest sum (a + b) {1 ≤ a ≤ N, 1 ≤ b ≤ N} such that a * b/(a + b) is an integer (i.e a + b divides a * b) and a != b.

Examples:

Input: N = 10
Output: 9
Explanation:
The numbers a = 3 and b = 6 leads to sum = 9 also 6 * 3 = 18 which is divisible by 6 + 3 = 9

Input: N = 20
Output: 27
Explanation:

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: The idea to solve this problem with the naive approach is to use the concept of nested loops. The following steps can be followed to compute the result:

1. Run a nested loop from 1 to N.
2. For every number a in the range [1, N], find another integer b such that a != b and (a + b) divides a * b.
3. If the condition is satisfied, the value of (a + b) is stored in a variable to keep the track of the maximum value obtained.
4. Finally, the maximum value of (a + b) is returned.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the largest value // of a + b satisfying the given condition #include using namespace std;    // Function to return the maximum sum of // a + b satisfying the given condition int getLargestSum(int N) {     // Initialize max_sum     int max_sum = 0;        // Consider all the possible pairs     for (int i = 1; i <= N; i++) {         for (int j = i + 1; j <= N; j++) {                // Check if the product is             // divisible by the sum             if (i * j % (i + j) == 0)                    // Storing the maximum sum                 // in the max_sum variable                 max_sum = max(max_sum, i + j);         }     }        // Return the max_sum value     return max_sum; }    // Driver code int main() {     int N = 25;        int max_sum = getLargestSum(N);        cout << max_sum << endl;     return 0; }

## Java

 // Java implementation to find the largest value // of a + b satisfying the given condition  import java.util.*;     class GFG{      // Function to return the maximum sum of // a + b satisfying the given condition static int getLargestSum(int N) {     // Initialize max_sum     int max_sum = 0;        // Consider all the possible pairs     for (int i = 1; i <= N; i++) {         for (int j = i + 1; j <= N; j++) {                // Check if the product is             // divisible by the sum             if (i * j % (i + j) == 0)                    // Storing the maximum sum                 // in the max_sum variable                 max_sum = Math.max(max_sum, i + j);         }     }        // Return the max_sum value     return max_sum; }    // Driver code public static void main(String[] args) {     int N = 25;        int max_sum = getLargestSum(N);        System.out.print(max_sum ); } }    // This code is contributed by shivanisinghss2110

## Python3

 # Python3 implementation to find the largest value # of a + b satisfying the given condition    # Function to return the maximum sum of # a + b satisfying the given condition def getLargestSum(N):     # Initialize max_sum     max_sum = 0        # Consider all the possible pairs     for i in range(1,N+1):         for j in range(i + 1, N + 1, 1):                # Check if the product is             # divisible by the sum             if (i * j % (i + j) == 0):                    # Storing the maximum sum                 # in the max_sum variable                 max_sum = max(max_sum, i + j)        # Return the max_sum value     return max_sum    # Driver code if __name__ == '__main__':     N = 25        max_sum = getLargestSum(N)     print(max_sum)    # This code is contributed by Surendra_Gangwar

## C#

 // C# implementation to find the largest value  // of a + b satisfying the given condition  using System;    class GFG{             // Function to return the maximum sum of      // a + b satisfying the given condition      static int getLargestSum(int N)      {          // Initialize max_sum          int max_sum = 0;                 // Consider all the possible pairs          for (int i = 1; i <= N; i++) {              for (int j = i + 1; j <= N; j++) {                         // Check if the product is                  // divisible by the sum                  if (i * j % (i + j) == 0)                             // Storing the maximum sum                      // in the max_sum variable                      max_sum = Math.Max(max_sum, i + j);              }          }                 // Return the max_sum value          return max_sum;      }             // Driver code      public static void Main(string[] args)      {          int N = 25;                 int max_sum = getLargestSum(N);                 Console.WriteLine(max_sum );      }  }     // This code is contributed by AnkitRai01

Output:

36

Time Complexity: Since the nested for loop runs (N * (N + 1)) / 2 times, the time complexity of the above solution is O(N2).

Efficient Approach: One observation which can be made is that if two numbers a and b exists such that their product is divisible by their sum then they are not relatively prime, i.e, their GCD is not one. This can be proved using Euclidean Algorithm.

Therefore, by using the above observation, the following steps can be followed to compute the result:

1. If a can be expressed as k * x and b can be expressed as k * y such that x and y are coprimes, then:
a + b = k(x + y)
a * b = k2

2. Now, upon dividing the above values:
(a * b) /(a + b) = k * ((x * y)/(x + y)).

3. Since it is known that x and y are coprime, (x * y) will not be divisible by (x + y). This means k must be divisible by x + y.
4. Therefore, k is the largest factor possible for which (k * x) and (k * y) remain less than N.
5. Clearly, the minimum value of k for which it is divisible by (x + y) is (x + y). This means that (x + y) * y &leq; N and (x + y) * x &leq; N.
6. Therefore, all the factors are checked from 1 to N0.5.

Below is the implementation of the above approach:

## C++

 // C++ implementation to  find the largest value // of a + b satisfying the given condition #include using namespace std;    // Function to return the maximum sum of // a + b satisfying the given condition int getLargestSum(int N) {     int max_sum = 0; // Initialize max_sum        // Consider all possible pairs and check     // the sum divides product property     for (int i = 1; i * i <= N; i++) {         for (int j = i + 1; j * j <= N; j++) {                // To find the largest factor k             int k = N / j;             int a = k * i;             int b = k * j;                // Check if the product is             // divisible by the sum             if (a <= N && b <= N                 && a * b % (a + b) == 0)                    // Storing the maximum sum                 // in the max_sum variable                 max_sum = max(max_sum, a + b);         }     }        // Return the max_sum value     return max_sum; }    // Driver code int main() {     int N = 25;     int max_sum = getLargestSum(N);        cout << max_sum << endl;     return 0; }

## Java

 // Java implementation to find the largest value // of a + b satisfying the given condition    class GFG{    // Function to return the maximum sum of // a + b satisfying the given condition static int getLargestSum(int N) {     // Initialize max_sum     int max_sum = 0;         // Consider all possible pairs and check     // the sum divides product property     for (int i = 1; i * i <= N; i++) {          for (int j = i + 1; j * j <= N; j++) {                  // To find the largest factor k               int k = N / j;               int a = k * i;               int b = k * j;                   // Check if the product is                // divisible by the sum                if (a <= N && b <= N &&                    a * b % (a + b) == 0)                       // Storing the maximum sum                    // in the max_sum variable                    max_sum = Math.max(max_sum, a + b);         }     }        // Return the max_sum value     return max_sum; }    // Driver code public static void main(String[] args) {     int N = 25;     int max_sum = getLargestSum(N);     System.out.print(max_sum + "\n"); } }    // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation to find the largest value  # of a + b satisfying the given condition     # Function to return the maximum sum of  # a + b satisfying the given condition  def getLargestSum(N) :         max_sum = 0; # Initialize max_sum         # Consider all possible pairs and check      # the sum divides product property      for i in range(1, int(N ** (1/2))+1) :         for j in range(i + 1, int(N ** (1/2)) + 1) :                            # To find the largest factor k             k = N // j;              a = k * i;             b = k * j;                            # Check if the product is             # divisible by the sum             if (a <= N and b <= N and a * b % (a + b) == 0) :                 # Storing the maximum sum                 # in the max_sum variable                  max_sum = max(max_sum, a + b);                        # Return the max_sum value     return max_sum;        # Driver code if __name__ == "__main__" :     N = 25;     max_sum = getLargestSum(N);     print(max_sum);    # This code is contributed by AnkitRai01

## C#

 // C# implementation to find the largest value // of a + b satisfying the given condition using System;    class GFG{    // Function to return the maximum sum of // a + b satisfying the given condition static int getLargestSum(int N) {            // Initialize max_sum     int max_sum = 0;         // Consider all possible pairs and check     // the sum divides product property     for(int i = 1; i * i <= N; i++)      {        for(int j = i + 1; j * j <= N; j++)        {           // To find the largest factor k           int k = N / j;           int a = k * i;           int b = k * j;                        // Check if the product is           // divisible by the sum           if (a <= N && b <= N &&               a * b % (a + b) == 0)                                // Storing the maximum sum               // in the max_sum variable               max_sum = Math.Max(max_sum, a + b);         }     }        // Return the max_sum value     return max_sum; }    // Driver code static public void Main(String[] args) {     int N = 25;     int max_sum = getLargestSum(N);            Console.Write(max_sum + "\n"); } }    // This code is contributed by gauravrajput1

Output:

36

Time Complexity: Though there are two loops, each loop runs for at most sqrt(N) time. Therefore, the overall time complexity O(N).

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