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# Count pairs of natural numbers with GCD equal to given number

Given three positive integer L, R, G. The task is to find the count of the pair (x,y) having GCD(x,y) = G and x, y lie between L and R.
Examples:

```Input : L = 1, R = 11, G = 5
Output : 3
(5, 5), (5, 10), (10, 5) are three pair having GCD equal to 5 and lie between 1 and 11.

Input : L = 1, R = 10, G = 7
Output : 1```

A simple solution is to go through all pairs in [L, R]. For every pair, find its GCD. If GCD is equal to g, then increment count. Finally return count.
An efficient solution is based on the fact that, for any positive integer pair (x, y) to have GCD equal to g, x and y should be divisible by g.
Observe, there will be at most (R – L)/g numbers between L and R which are divisible by g.
So we find numbers between L and R which are divisible by g. For this, we start from ceil(L/g) * g and with increment by g at each step while it doesn’t exceed R, count numbers having GCD equal to 1.
Also,

`ceil(L/g) * g = floor((L + g - 1) / g) * g.`

Below is the implementation of above idea :

## C++

 `// C++ program to count pair in range of natural``// number having GCD equal to given number.``#include ``using` `namespace` `std;` `// Return the GCD of two numbers.``int` `gcd(``int` `a, ``int` `b)``{``    ``return` `b ? gcd(b, a % b) : a;``}` `// Return the count of pairs having GCD equal to g.``int` `countGCD(``int` `L, ``int` `R, ``int` `g)``{``    ``// Setting the value of L, R.``    ``L = (L + g - 1) / g;``    ``R = R/ g;` `    ``// For each possible pair check if GCD is 1.``    ``int` `ans = 0;``    ``for` `(``int` `i = L; i <= R; i++)``        ``for` `(``int` `j = L; j <= R; j++)``            ``if` `(gcd(i, j) == 1)``                ``ans++;` `    ``return` `ans;``}` `// Driven Program``int` `main()``{``    ``int` `L = 1, R = 11, g = 5;``    ``cout << countGCD(L, R, g) << endl;``    ``return` `0;``}`

## Java

 `// Java program to count pair in``// range of natural number having``// GCD equal to given number.``import` `java.util.*;` `class` `GFG {``    ` `// Return the GCD of two numbers.``static` `int` `gcd(``int` `a, ``int` `b)``{``    ``return` `b > ``0` `? gcd(b, a % b) : a;``}` `// Return the count of pairs``// having GCD equal to g.``static` `int` `countGCD(``int` `L, ``int` `R, ``int` `g) {``    ` `    ``// Setting the value of L, R.``    ``L = (L + g - ``1``) / g;``    ``R = R / g;` `    ``// For each possible pair check if GCD is 1.``    ``int` `ans = ``0``;``    ``for` `(``int` `i = L; i <= R; i++)``    ``for` `(``int` `j = L; j <= R; j++)``        ``if` `(gcd(i, j) == ``1``)``        ``ans++;` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args) {``    ` `    ``int` `L = ``1``, R = ``11``, g = ``5``;``    ``System.out.println(countGCD(L, R, g));``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python program to count``# pair in range of natural``# number having GCD equal``# to given number.` `# Return the GCD of two numbers.``def` `gcd(a,b):` `    ``return` `gcd(b, a ``%` `b) ``if` `b>``0` `else` `a` ` ` `# Return the count of pairs``# having GCD equal to g.``def` `countGCD(L,R,g):` `    ``# Setting the value of L, R.``    ``L ``=` `(L ``+` `g ``-` `1``) ``/``/` `g``    ``R ``=` `R``/``/` `g`` ` `    ``# For each possible pair``    ``# check if GCD is 1.``    ``ans ``=` `0``    ``for` `i ``in` `range``(L,R``+``1``):``        ``for` `j ``in` `range``(L,R``+``1``):``            ``if` `(gcd(i, j) ``=``=` `1``):``                ``ans``=``ans ``+``1`` ` `    ``return` `ans` `# Driver code` `L ``=` `1``R ``=` `11``g ``=` `5` `print``(countGCD(L, R, g))` `# This code is contributed``# by Anant Agarwal.`

## C#

 `// C# program to count pair in``// range of natural number having``// GCD equal to given number.``using` `System;` `class` `GFG {``    ` `// Return the GCD of two numbers.``static` `int` `gcd(``int` `a, ``int` `b)``{``    ``return` `b > 0 ? gcd(b, a % b) : a;``}` `// Return the count of pairs``// having GCD equal to g.``static` `int` `countGCD(``int` `L, ``int` `R,``                    ``int` `g)``{``    ` `    ``// Setting the value of L, R.``    ``L = (L + g - 1) / g;``    ``R = R / g;` `    ``// For each possible pair``    ``// check if GCD is 1.``    ``int` `ans = 0;``    ``for` `(``int` `i = L; i <= R; i++)``    ``for` `(``int` `j = L; j <= R; j++)``        ``if` `(gcd(i, j) == 1)``        ``ans++;` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main()``{``    ` `    ``int` `L = 1, R = 11, g = 5;``    ``Console.WriteLine(countGCD(L, R, g));``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output:

`3`

Time Complexity : O((r-l)*(r-l)*log(min(k))) where l and r are lower limit, upper limit and k is the number between l and r.

Space Complexity : O(logk)

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.