Count pairs of natural numbers with GCD equal to given number

Last Updated : 12 Sep, 2023

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# 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:

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)

Count pairs of natural numbers with GCD equal to given number

kartik

Improve

Article Tags :

Practice Tags :

Mathematical

Count pairs of natural numbers with GCD equal to given number

Similar Reads