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Count numbers in range 1 to N which are divisible by X but not by Y
• Last Updated : 20 Mar, 2019

Given two positive integers X and Y, the task is to count the total numbers in range 1 to N which are divisible by X but not Y.

Examples:

Input: x = 2, Y = 3, N = 10
Output: 4
Numbers divisible by 2 but not 3 are : 2, 4, 8, 10

Input : X = 2, Y = 4, N = 20
Output : 5
Numbers divisible by 2 but not 4 are : 2, 6, 10, 14, 18

## Recommended: Please try your approach on {IDE}first, before moving on to the solution.

A Simple Solution is to count numbers divisible by X but not Y is to loop through 1 to N and counting such number which is divisible by X but not Y.

Approach

1. For every number in range 1 to N, Increment count if the number is divisible by X but not by Y.
2. Print the count.

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;`` ` `// Function to count total numbers divisible by``// x but not y in range 1 to N``int` `countNumbers(``int` `X, ``int` `Y, ``int` `N)``{``    ``int` `count = 0;``    ``for` `(``int` `i = 1; i <= N; i++) {``        ``// Check if Number is divisible``        ``// by x but not Y``        ``// if yes, Increment count``        ``if` `((i % X == 0) && (i % Y != 0))``            ``count++;``    ``}``    ``return` `count;``}`` ` `// Driver Code``int` `main()``{`` ` `    ``int` `X = 2, Y = 3, N = 10;``    ``cout << countNumbers(X, Y, N);``    ``return` `0;``}`

## Java

 `// Java implementation of above approach`` ` `class` `GFG {`` ` `    ``// Function to count total numbers divisible by``    ``// x but not y in range 1 to N``    ``static` `int` `countNumbers(``int` `X, ``int` `Y, ``int` `N)``    ``{``        ``int` `count = ``0``;``        ``for` `(``int` `i = ``1``; i <= N; i++) {``            ``// Check if Number is divisible``            ``// by x but not Y``            ``// if yes, Increment count``            ``if` `((i % X == ``0``) && (i % Y != ``0``))``                ``count++;``        ``}``        ``return` `count;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{`` ` `        ``int` `X = ``2``, Y = ``3``, N = ``10``;``        ``System.out.println(countNumbers(X, Y, N));``    ``}``}`

## Python3

 `# Python3 implementation of above approach `` ` `# Function to count total numbers divisible ``# by x but not y in range 1 to N ``def` `countNumbers(X, Y, N): `` ` `    ``count ``=` `0``; ``    ``for` `i ``in` `range``(``1``, N ``+` `1``):``         ` `        ``# Check if Number is divisible ``        ``# by x but not Y ``        ``# if yes, Increment count ``        ``if` `((i ``%` `X ``=``=` `0``) ``and` `(i ``%` `Y !``=` `0``)): ``            ``count ``+``=` `1``; `` ` `    ``return` `count; `` ` `# Driver Code ``X ``=` `2``;``Y ``=` `3``;``N ``=` `10``; ``print``(countNumbers(X, Y, N)); ``     ` `# This code is contributed by mits`

## C#

 `// C# implementation of the above approach``using` `System;``class` `GFG {`` ` `    ``// Function to count total numbers divisible by``    ``// x but not y in range 1 to N``    ``static` `int` `countNumbers(``int` `X, ``int` `Y, ``int` `N)``    ``{``        ``int` `count = 0;``        ``for` `(``int` `i = 1; i <= N; i++) {``            ``// Check if Number is divisible``            ``// by x but not Y``            ``// if yes, Increment count``            ``if` `((i % X == 0) && (i % Y != 0))``                ``count++;``        ``}``        ``return` `count;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{`` ` `        ``int` `X = 2, Y = 3, N = 10;``        ``Console.WriteLine(countNumbers(X, Y, N));``    ``}``}`

## PHP

 ``
Output:
```4
```

Time Complexity : O(N)

Efficient solution:

1. In range 1 to N, find total numbers divisible by X and total numbers divisible by Y.
2. Also, Find total numbers divisible by either X or Y
3. Calculate total number divisible by X but not Y as
(total number divisible by X or Y) – (total number divisible by Y)

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;`` ` `// Function to count total numbers divisible by``// x but not y in range 1 to N``int` `countNumbers(``int` `X, ``int` `Y, ``int` `N)``{`` ` `    ``// Count total number divisible by X``    ``int` `divisibleByX = N / X;`` ` `    ``// Count total number divisible by Y``    ``int` `divisibleByY = N / Y;`` ` `    ``// Count total number divisible by either X or Y``    ``int` `LCM = (X * Y) / __gcd(X, Y);``    ``int` `divisibleByLCM = N / LCM;``    ``int` `divisibleByXorY = divisibleByX + divisibleByY ``                                     ``- divisibleByLCM;`` ` `    ``// Count total numbers divisible by X but not Y``    ``int` `divisibleByXnotY = divisibleByXorY ``                                       ``- divisibleByY;`` ` `    ``return` `divisibleByXnotY;``}`` ` `// Driver Code``int` `main()``{`` ` `    ``int` `X = 2, Y = 3, N = 10;``    ``cout << countNumbers(X, Y, N);``    ``return` `0;``}`

## Java

 `// Java implementation of above approach`` ` `class` `GFG {`` ` `    ``// Function to calculate GCD`` ` `    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(b == ``0``)``            ``return` `a;``        ``return` `gcd(b, a % b);``    ``}`` ` `    ``// Function to count total numbers divisible by``    ``// x but not y in range 1 to N`` ` `    ``static` `int` `countNumbers(``int` `X, ``int` `Y, ``int` `N)``    ``{`` ` `        ``// Count total number divisible by X``        ``int` `divisibleByX = N / X;`` ` `        ``// Count total number divisible by Y``        ``int` `divisibleByY = N / Y;`` ` `        ``// Count total number divisible by either X or Y``        ``int` `LCM = (X * Y) / gcd(X, Y);``        ``int` `divisibleByLCM = N / LCM;``        ``int` `divisibleByXorY = divisibleByX + divisibleByY``                              ``- divisibleByLCM;`` ` `        ``// Count total number divisible by X but not Y``        ``int` `divisibleByXnotY = divisibleByXorY ``                                          ``- divisibleByY;`` ` `        ``return` `divisibleByXnotY;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{`` ` `        ``int` `X = ``2``, Y = ``3``, N = ``10``;``        ``System.out.println(countNumbers(X, Y, N));``    ``}``}`

## Python3

 `# Python 3 implementation of above approach``from` `math ``import` `gcd`` ` `# Function to count total numbers divisible ``# by x but not y in range 1 to N``def` `countNumbers(X, Y, N):``     ` `    ``# Count total number divisible by X``    ``divisibleByX ``=` `int``(N ``/` `X)`` ` `    ``# Count total number divisible by Y``    ``divisibleByY ``=` `int``(N ``/` `Y)`` ` `    ``# Count total number divisible ``    ``# by either X or Y``    ``LCM ``=` `int``((X ``*` `Y) ``/` `gcd(X, Y))``    ``divisibleByLCM ``=` `int``(N ``/` `LCM)``    ``divisibleByXorY ``=` `(divisibleByX ``+` `                       ``divisibleByY ``-` `                       ``divisibleByLCM)`` ` `    ``# Count total numbers divisible by ``    ``# X but not Y``    ``divisibleByXnotY ``=` `(divisibleByXorY ``-` `                        ``divisibleByY)`` ` `    ``return` `divisibleByXnotY`` ` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``X ``=` `2``    ``Y ``=` `3``    ``N ``=` `10``    ``print``(countNumbers(X, Y, N))`` ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of above approach`` ` `using` `System;``class` `GFG {`` ` `    ``// Function to calculate GCD``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(b == 0)``            ``return` `a;``        ``return` `gcd(b, a % b);``    ``}`` ` `    ``// Function to count total numbers divisible by``    ``// x but not y in range 1 to N``    ``static` `int` `countNumbers(``int` `X, ``int` `Y, ``int` `N)``    ``{`` ` `        ``// Count total number divisible by X``        ``int` `divisibleByX = N / X;`` ` `        ``// Count total number divisible by Y``        ``int` `divisibleByY = N / Y;`` ` `        ``// Count total number divisible by either X or Y``        ``int` `LCM = (X * Y) / gcd(X, Y);``        ``int` `divisibleByLCM = N / LCM;``        ``int` `divisibleByXorY = divisibleByX + divisibleByY ``                                        ``- divisibleByLCM;`` ` `        ``// Count total number divisible by X but not Y``        ``int` `divisibleByXnotY = divisibleByXorY ``                                          ``- divisibleByY;`` ` `        ``return` `divisibleByXnotY;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{`` ` `        ``int` `X = 2, Y = 3, N = 10;``        ``Console.WriteLine(countNumbers(X, Y, N));``    ``}``}`

## PHP

 ` ``\$b``) ``        ``return` `__gcd( ``\$a` `- ``\$b` `, ``\$b` `); `` ` `    ``return` `__gcd( ``\$a` `, ``\$b` `- ``\$a` `); ``} `` ` `// Function to count total numbers divisible ``// by x but not y in range 1 to N``function` `countNumbers(``\$X``, ``\$Y``, ``\$N``)``{`` ` `    ``// Count total number divisible by X``    ``\$divisibleByX` `= ``\$N` `/ ``\$X``;`` ` `    ``// Count total number divisible by Y``    ``\$divisibleByY` `= ``\$N` `/``\$Y``;`` ` `    ``// Count total number divisible by either X or Y``    ``\$LCM` `= (``\$X` `* ``\$Y``) / __gcd(``\$X``, ``\$Y``);``    ``\$divisibleByLCM` `= ``\$N` `/ ``\$LCM``;``    ``\$divisibleByXorY` `= ``\$divisibleByX` `+ ``\$divisibleByY` `- ``                                       ``\$divisibleByLCM``;`` ` `    ``// Count total numbers divisible by X but not Y``    ``\$divisibleByXnotY` `= ``\$divisibleByXorY` `- ``                        ``\$divisibleByY``;`` ` `    ``return` `ceil``(``\$divisibleByXnotY``);``}`` ` `// Driver Code``\$X` `= 2;``\$Y` `= 3;``\$N` `= 10;``echo` `countNumbers(``\$X``, ``\$Y``, ``\$N``);`` ` `// This is code contrubted by inder_verma``?>`
Output:
```4
```

Time Complexity: O(1)

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