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

Given a number N, 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 

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 N 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++

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// C++ program to find distinct integers
// ontained by lcm(x, num)/x
#include <cmath>
#include <iostream>
  
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;
}

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Java

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// 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)); 
  
  
    }
}

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Python 3

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# 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

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C#

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// 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

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PHP

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<?php
// PHP program to find distinct 
// integers ontained by lcm(x, num)/x
  
// Function to count the number 
// of distinct integers ontained
// by lcm(x, num)/x
function numberOfDistinct($n)
{
    $ans = 0;
  
    // iterate to count the 
    // number of factors
    for ($i = 1; $i <= sqrt($n); $i++)
    {
        if ($n % $i == 0) 
        {
            $ans++;
            if (($n / $i) != $i)
                $ans++;
        }
    }
  
    return $ans;
}
  
// Driver Code
$n = 3;
echo numberOfDistinct($n);
  
// This code is contributed by ajit
?>

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Output:

2

Time Complexity: O(sqrt(n))



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Striver(underscore)79 at Codechef and codeforces D

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Improved By : jit_t, chitranayal