# 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. So answer is 3. 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 <bits/stdc++.h>` `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

`<?php` `// PHP program to count pair` `// in range of natural number` `// having GCD equal to given number.` `// Return the GCD of two numbers.` `function` `gcd( ` `$a` `, ` `$b` `)` `{` ` ` `return` `$b` `? gcd(` `$b` `, ` `$a` `% ` `$b` `) : ` `$a` `;` `}` `// Return the count of pairs` `// having GCD equal to g.` `function` `countGCD( ` `$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` `(` `$i` `= ` `$L` `; ` `$i` `<= ` `$R` `; ` `$i` `++)` ` ` `for` `(` `$j` `= ` `$L` `; ` `$j` `<= ` `$R` `; ` `$j` `++)` ` ` `if` `(gcd(` `$i` `, ` `$j` `) == 1)` ` ` `$ans` `++;` ` ` `return` `$ans` `;` `}` ` ` `// Driver Code` ` ` `$L` `= 1;` ` ` `$R` `= 11;` ` ` `$g` `= 5;` ` ` `echo` `countGCD(` `$L` `, ` `$R` `, ` `$g` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` ` ` `// Javascript program to count pair in` ` ` `// range of natural number having` ` ` `// GCD equal to given number.` ` ` ` ` `// Return the GCD of two numbers.` ` ` `function` `gcd(a, b)` ` ` `{` ` ` `return` `b > 0 ? gcd(b, a % b) : a;` ` ` `}` ` ` `// Return the count of pairs` ` ` `// having GCD equal to g.` ` ` `function` `countGCD(L, R, g)` ` ` `{` ` ` `// Setting the value of L, R.` ` ` `L = parseInt((L + g - 1) / g, 10);` ` ` `R = parseInt(R / g, 10);` ` ` `// For each possible pair` ` ` `// check if GCD is 1.` ` ` `let ans = 0;` ` ` `for` `(let i = L; i <= R; i++)` ` ` `for` `(let j = L; j <= R; j++)` ` ` `if` `(gcd(i, j) == 1)` ` ` `ans++;` ` ` `return` `ans;` ` ` `}` ` ` ` ` `let L = 1, R = 11, g = 5;` ` ` `document.write(countGCD(L, R, g));` ` ` `</script>` |

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.

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