# 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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