# Count pairs of numbers from 1 to N with Product divisible by their Sum

Given a number . The task is to count pairs (x, y) such that x*y is divisible by (x+y) and the condition 1 <= x < y < N holds true.

Examples:

Input : N = 6
Output : 1
Explanation: The only pair is (3, 6) which satisfies
all of the given condition, 3<6 and 18%9=0.

Input : N = 15
Output : 4

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

The basic approach is to iterate using two loops carefully maintaining the given condition 1 <= x < y < N and generate all possible valid pairs and count such pairs for which the product of their values is divisible by sum.

Below is the implementation of the above approach:

## C++

 // C++ program to count pairs of numbers // from 1 to N with Product divisible // by their Sum    #include using namespace std;    // Function to count pairs int countPairs(int n) {     // variable to store count     int count = 0;        // Generate all possible pairs such that     // 1 <= x < y < n     for (int x = 1; x < n; x++) {         for (int y = x + 1; y <= n; y++) {             if ((y * x) % (y + x) == 0)                 count++;         }     }        return count; }    // Driver code int main() {     int n = 15;        cout << countPairs(n);        return 0; }

## Java

 // Java program to count pairs of numbers // from 1 to N with Product divisible // by their Sum    import java.io.*;    class GFG {           // Function to count pairs static int countPairs(int n) {     // variable to store count     int count = 0;        // Generate all possible pairs such that     // 1 <= x < y < n     for (int x = 1; x < n; x++) {         for (int y = x + 1; y <= n; y++) {             if ((y * x) % (y + x) == 0)                 count++;         }     }        return count; }    // Driver code        public static void main (String[] args) {             int n = 15;        System.out.println(countPairs(n));     } } // This code is contributed by anuj_67..

## Python3

 # Python 3 program to count pairs of numbers # from 1 to N with Product divisible # by their Sum    # Function to count pairs def countPairs(n):            # variable to store count     count = 0            # Generate all possible pairs such that     # 1 <= x < y < n     for x in range(1, n):         for y in range(x + 1, n + 1):             if ((y * x) % (y + x) == 0):                 count += 1        return count    # Driver code n = 15 print(countPairs(n))    # This code is contributed  # by PrinciRaj1992

## C#

 // C# program to count pairs of numbers // from 1 to N with Product divisible // by their Sum using System;    class GFG  {    // Function to count pairs static int countPairs(int n) {     // variable to store count     int count = 0;        // Generate all possible pairs      // such that 1 <= x < y < n     for (int x = 1; x < n; x++)      {         for (int y = x + 1; y <= n; y++)         {             if ((y * x) % (y + x) == 0)                 count++;         }     }        return count; }    // Driver code public static void Main () {     int n = 15;        Console.WriteLine(countPairs(n)); } }    // This code is contributed by anuj_67

## PHP



Output:

4

Time Complexity : O(N2)

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Improved By : vt_m, jit_t, princiraj1992

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