# Count all possible values of K less than Y such that GCD(X, Y) = GCD(X+K, Y)

Given two integers X and Y, the task is to find the number of integers, K, such that gcd(X, Y) is equal to gcd(X+K, Y), where 0 < K <Y.

Examples:

Input: X = 3, Y = 15
Output: 4
Explanation: All possible values of K are {0, 3, 6, 9} for which GCD(X, Y) = GCD(X + K, Y).

Input: X = 2, Y = 12
Output: 2
Explanation: All possible values of K are {0, 8}.

Naive Approach: The simplest approach to solve the problem is to iterate over the range [0, Y – 1]and for each value of i, check if GCD(X + i, Y) is equal to GCD(X, Y) or not.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to calculate` `// GCD of two integers` `int` `gcd(``int` `a, ``int` `b)` `{` `    ``if` `(b == 0)` `        ``return` `a;`   `    ``return` `gcd(b, a % b);` `}`   `// Function to count possible` `// values of K` `int` `calculateK(``int` `x, ``int` `y)` `{` `    ``int` `count = 0;` `    ``int` `gcd_xy = gcd(x, y);` `    ``for` `(``int` `i = 0; i < y; i++) {`   `        ``// If required condition` `        ``// is satisfied` `        ``if` `(gcd(x + i, y) == gcd_xy)`   `            ``// Increase count` `            ``count++;` `    ``}`   `    ``return` `count;` `}`   `// Driver Code` `int` `main()` `{`   `    ``// Given X and y` `    ``int` `x = 3, y = 15;`   `    ``cout << calculateK(x, y) << endl;` `}`

## Java

 `// Java program for the above approach ` `import` `java.util.*;` `class` `GFG` `{` `      `  `// Function to calculate ` `// GCD of two integers ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == ``0``) ` `        ``return` `a;   ` `    ``return` `gcd(b, a % b); ` `} ` `  `  `// Function to count possible ` `// values of K ` `static` `int` `calculateK(``int` `x, ``int` `y) ` `{ ` `    ``int` `count = ``0``; ` `    ``int` `gcd_xy = gcd(x, y); ` `    ``for` `(``int` `i = ``0``; i < y; i++) ` `    ``{ ` `  `  `        ``// If required condition ` `        ``// is satisfied ` `        ``if` `(gcd(x + i, y) == gcd_xy) ` `  `  `            ``// Increase count ` `            ``count++; ` `    ``}   ` `    ``return` `count; ` `} ` `  `  `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``// Given X and y ` `    ``int` `x = ``3``, y = ``15``;  ` `    ``System.out.print(calculateK(x, y)); ` `}` `}`   `// This code is contributed by sanjoy_62`

## Python3

 `# Python3 program for the above approach`   `# Function to calculate` `# GCD of two integers` `def` `gcd(a, b):` `    `  `    ``if` `(b ``=``=` `0``):` `        ``return` `a`   `    ``return` `gcd(b, a ``%` `b)`   `# Function to count possible` `# values of K` `def` `calculateK(x, y):` `    `  `    ``count ``=` `0` `    ``gcd_xy ``=` `gcd(x, y)`   `    ``for` `i ``in` `range``(y):` `        `  `        ``# If required condition` `        ``# is satisfied` `        ``if` `(gcd(x ``+` `i, y) ``=``=` `gcd_xy):` `            `  `            ``# Increase count` `            ``count ``+``=` `1`   `    ``return` `count`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# Given X and y` `    ``x ``=` `3` `    ``y ``=` `15`   `    ``print` `(calculateK(x, y))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach ` `using` `System;`   `class` `GFG{` `      `  `// Function to calculate ` `// GCD of two integers ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == 0) ` `        ``return` `a;  ` `        `  `    ``return` `gcd(b, a % b); ` `} ` `  `  `// Function to count possible ` `// values of K ` `static` `int` `calculateK(``int` `x, ``int` `y) ` `{ ` `    ``int` `count = 0; ` `    ``int` `gcd_xy = gcd(x, y); ` `    `  `    ``for``(``int` `i = 0; i < y; i++) ` `    ``{ ` `        `  `        ``// If required condition ` `        ``// is satisfied ` `        ``if` `(gcd(x + i, y) == gcd_xy) ` `        `  `            ``// Increase count ` `            ``count++; ` `    ``}   ` `    ``return` `count; ` `} ` `  `  `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    `  `    ``// Given X and y ` `    ``int` `x = 3, y = 15;  ` `    `  `    ``Console.Write(calculateK(x, y)); ` `}` `}`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`4`

Time Complexity: O(YlogY)
Auxiliary Space: O(1)

Efficient Approach: The idea is to use the concept of Euler’s totient function. Follow the steps below to solve the problem:

• Calculate the gcd of X and Y and store it in a variable g.
•  Initialize a variable n with Y/g.
•  Now, find the totient function for n which will be the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the gcd of a and b` `int` `gcd(``int` `a, ``int` `b)` `{`   `    ``if` `(b == 0)` `        ``return` `a;`   `    ``return` `gcd(b, a % b);` `}`   `// Function to find the number of Ks` `int` `calculateK(``int` `x, ``int` `y)` `{`   `    ``// Find gcd` `    ``int` `g = gcd(x, y);` `    ``int` `n = y / g;` `    ``int` `res = n;`   `    ``// Calculating value of totient` `    ``// function for n` `    ``for` `(``int` `i = 2; i * i <= n; i++) {` `        ``if` `(n % i == 0) {` `            ``res -= (res / i);` `            ``while` `(n % i == 0)` `                ``n /= i;` `        ``}` `    ``}` `    ``if` `(n != 1)` `        ``res -= (res / n);` `    ``return` `res;` `}`   `// Driver Code` `int` `main()` `{`   `    ``// Given X and Y` `    ``int` `x = 3, y = 15;`   `    ``cout << calculateK(x, y) << endl;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG` `{`   `// Function to find the gcd of a and b` `static` `int` `gcd(``int` `a, ``int` `b)` `{`   `    ``if` `(b == ``0``)` `        ``return` `a;` `    ``return` `gcd(b, a % b);` `}`   `// Function to find the number of Ks` `static` `int` `calculateK(``int` `x, ``int` `y)` `{`   `    ``// Find gcd` `    ``int` `g = gcd(x, y);` `    ``int` `n = y / g;` `    ``int` `res = n;`   `    ``// Calculating value of totient` `    ``// function for n` `    ``for` `(``int` `i = ``2``; i * i <= n; i++)` `    ``{` `        ``if` `(n % i == ``0``) ` `        ``{` `            ``res -= (res / i);` `            ``while` `(n % i == ``0``)` `                ``n /= i;` `        ``}` `    ``}` `    ``if` `(n != ``1``)` `        ``res -= (res / n);` `    ``return` `res;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{`   `    ``// Given X and Y` `    ``int` `x = ``3``, y = ``15``;` `    ``System.out.print(calculateK(x, y) +``"\n"``);` `}` `}`   `// This code is contributed by shikhasingrajput`

## Python3

 `# Python 3 program for the above approach`   `# Function to find the gcd of a and b` `def` `gcd(a, b):` `    ``if` `(b ``=``=` `0``):` `        ``return` `a` `    ``return` `gcd(b, a ``%` `b)`   `# Function to find the number of Ks` `def` `calculateK(x, y):`   `    ``# Find gcd` `    ``g ``=` `gcd(x, y)` `    ``n ``=` `y ``/``/` `g` `    ``res ``=` `n`   `    ``# Calculating value of totient` `    ``# function for n` `    ``i ``=` `2` `    ``while` `i ``*` `i <``=` `n:` `        ``if` `(n ``%` `i ``=``=` `0``):` `            ``res ``-``=` `(res ``/``/` `i)` `            ``while` `(n ``%` `i ``=``=` `0``):` `                ``n ``/``/``=` `i` `        ``i ``+``=` `1` `    ``if` `(n !``=` `1``):` `        ``res ``-``=` `(res ``/``/` `n)` `    ``return` `res`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``# Given X and Y` `    ``x ``=` `3` `    ``y ``=` `15`   `    ``print``(calculateK(x, y))` `    `  `    ``# This code is contributed by chitranayal.`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{`   `// Function to find the gcd of a and b` `static` `int` `gcd(``int` `a, ``int` `b)` `{` `    ``if` `(b == 0)` `        ``return` `a;` `        `  `    ``return` `gcd(b, a % b);` `}`   `// Function to find the number of Ks` `static` `int` `calculateK(``int` `x, ``int` `y)` `{` `    `  `    ``// Find gcd` `    ``int` `g = gcd(x, y);` `    ``int` `n = y / g;` `    ``int` `res = n;`   `    ``// Calculating value of totient` `    ``// function for n` `    ``for``(``int` `i = 2; i * i <= n; i++)` `    ``{` `        ``if` `(n % i == 0) ` `        ``{` `            ``res -= (res / i);` `            `  `            ``while` `(n % i == 0)` `                ``n /= i;` `        ``}` `    ``}` `    ``if` `(n != 1)` `        ``res -= (res / n);` `        `  `    ``return` `res;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    `  `    ``// Given X and Y` `    ``int` `x = 3, y = 15;` `    `  `    ``Console.Write(calculateK(x, y) + ``"\n"``);` `}` `}`   `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output:

`4`

Time Complexity: O(log(min(X, Y)) + ?N) where N is Y/gcd(X, Y).
Auxiliary Space: O(1)

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