# 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

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 pairsint 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 codeint 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 pairsstatic 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 pairsdef 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 coden = 15print(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 Sumusing System; class GFG { // Function to count pairsstatic 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 codepublic static void Main (){    int n = 15;     Console.WriteLine(countPairs(n));}} // This code is contributed by anuj_67

## PHP



## Javascript



Output:
4

Time Complexity : O(N2)

Auxiliary Space: O(1)

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