# 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; ` `} ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` `?> ` |

*chevron_right*

*filter_none*

Output:

3

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Number of pairs from the first N natural numbers whose sum is divisible by K
- Count the numbers < N which have equal number of divisors as K
- Count pairs from two arrays having sum equal to K
- Count pairs with set bits sum equal to K
- Count natural numbers whose factorials are divisible by x but not y
- Count of Numbers in Range where first digit is equal to last digit of the number
- Count pairs of numbers from 1 to N with Product divisible by their Sum
- Count ordered pairs of positive numbers such that their sum is S and XOR is K
- Count numbers whose XOR with N is equal to OR with N
- Find if given number is sum of first n natural numbers
- Count numbers whose difference with N is equal to XOR with N
- Count different numbers that can be generated such that there digits sum is equal to 'n'
- Sum of first N natural numbers by taking powers of 2 as negative number
- Number of distinct prime factors of first n natural numbers
- Find the number of sub arrays in the permutation of first N natural numbers such that their median is M