# Number of distinct integers obtained by lcm(X, N)/X

Given a number , find the number of distinct integers obtained by lcm(X, N)/X where X can be any positive number.

Examples:

```Input: N = 2
Output: 2
if X is 1, then lcm(1, 2)/1 is 2/1=2.
if X is 2, then lcm(2, 2)/2 is 2/2=1.
For any X greater than 2 we cannot
obtain a distinct integer.

Input: N = 3
Output: 2
```

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

It is known that lcm(x, y) = x*y/gcd(x, y).

Therefore,

```lcm(X, N) = X*N/gcd(X, N)
or, lcm(X, N)/X = N/gcd(X, N)
```

So only the distinct factors of can be the distinct integers possible. Hence count the number of distinct factors of N including 1 and N itself, which is the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to find distinct integers ` `// ontained by lcm(x, num)/x ` `#include ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to count the number of distinct ` `// integers ontained by lcm(x, num)/x ` `int` `numberOfDistinct(``int` `n) ` `{ ` `    ``int` `ans = 0; ` ` `  `    ``// iterate to count the number of factors ` `    ``for` `(``int` `i = 1; i <= ``sqrt``(n); i++) { ` `        ``if` `(n % i == 0) { ` `            ``ans++; ` `            ``if` `((n / i) != i) ` `                ``ans++; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` ` `  `    ``cout << numberOfDistinct(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java  program to find distinct integers  ` `// ontained by lcm(x, num)/x  ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `// Function to count the number of distinct  ` `// integers ontained by lcm(x, num)/x  ` `static` `int` `numberOfDistinct(``int` `n)  ` `{  ` `    ``int` `ans = ``0``;  ` ` `  `    ``// iterate to count the number of factors  ` `    ``for` `(``int` `i = ``1``; i <= Math.sqrt(n); i++) {  ` `        ``if` `(n % i == ``0``) {  ` `            ``ans++;  ` `            ``if` `((n / i) != i)  ` `                ``ans++;  ` `        ``}  ` `    ``}  ` ` `  `    ``return` `ans;  ` `}  ` ` `  `// Driver Code  ` `    ``public` `static` `void` `main (String[] args) { ` `        ``int` `n = ``3``;  ` ` `  `        ``System.out.println (numberOfDistinct(n));  ` ` `  ` `  `    ``} ` `} `

## Python 3

 `# Python 3 program to find distinct integers ` `# ontained by lcm(x, num)/x ` ` `  `import` `math ` ` `  `# Function to count the number of distinct ` `# integers ontained by lcm(x, num)/x ` `def` `numberOfDistinct(n): ` `    ``ans ``=` `0` ` `  `    ``# iterate to count the number of factors ` `    ``for` `i ``in` `range``( ``1``, ``int``(math.sqrt(n))``+``1``): ` `        ``if` `(n ``%` `i ``=``=` `0``) : ` `            ``ans ``+``=` `1` `            ``if` `((n ``/``/` `i) !``=` `i): ` `                ``ans ``+``=` `1` `    ``return` `ans ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``n ``=` `3` ` `  `    ``print``(numberOfDistinct(n)) ` ` `  `# This code is contributed by ` `# ChitraNayal `

## C#

 `// C# program to find distinct integers  ` `// ontained by lcm(x, num)/x  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to count the number  ` `// of distinct integers ontained ` `// by lcm(x, num)/x  ` `static` `int` `numberOfDistinct(``int` `n)  ` `{  ` `    ``int` `ans = 0;  ` ` `  `    ``// iterate to count the number ` `    ``// of factors  ` `    ``for` `(``int` `i = 1; i <= Math.Sqrt(n); i++)  ` `    ``{  ` `        ``if` `(n % i == 0) ` `        ``{  ` `            ``ans++;  ` `            ``if` `((n / i) != i)  ` `                ``ans++;  ` `        ``}  ` `    ``}  ` ` `  `    ``return` `ans;  ` `}  ` ` `  `// Driver Code ` `static` `public` `void` `Main () ` `{ ` `    ``int` `n = 3;  ` `    ``Console.WriteLine(numberOfDistinct(n));  ` `} ` `} ` ` `  `// This code is contributed by ajit `

## PHP

 ` `

Output:

```2
```

Time Complexity: O(sqrt(n))

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