Count ordered pairs of numbers with a given LCM

Given an integer N, the task is to count the total number of ordered pairs such that the LCM of each pair is equal to N.

Examples:

Input: N = 6
Output:
Explanation: 
Pairs with LCM equal to N(= 6) are {(1, 6), (2, 6), (2, 3), (3, 6), (6, 6), (6, 3), (3, 2), (6, 2), (6, 1)} 
Therefore, the output is 9.

Input: N = 36
Output: 25

 

Approach: The problem can be solved based on the following observations:



Consider an ordered pair(X, Y). 
X = P1a1 * P2a2 * P3a3 *…..* Pnan 
Y = P1b1 * P2b2 * P3b3 *…..* Pnbn
Here, P1, P2, ….., Pn are prime factors of X and Y. 
LCM(X, Y) = P1max(a1, b1)  * P2max(a2, b2) *……….*Pnmax(an, bn)
Therefore, LCM(X, Y) = N = P1m1 * P2m2 * P3m3 *…..* Pnmn

Therefore, total number of ordered pairs (X, Y) 
= [{(m1 + 1)2 – m12} * {(m2 + 1)2 – m22} * ……* {(mn + 1)2 – mn2} ]
= (2*m1+1) * (2*m2+1) * (2*m3+1) * ……..* (2*mn+1).

Follow the steps below to solve the problem:

  1. Initialize a variable, say, countPower, to store the power of all prime factors of N.
  2. Calculate the power of all prime factors of N.
  3. Finally, print the count of ordered pairs(X, Y) using the aforementioned formula.

Below is the implementation of the above approach:
 

C++

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// C++ program to implement
// the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the number of
// ordered pairs with given LCM
int CtOrderedPairs(int N)
{
    // Stores count of
    // ordered pairs
    int res = 1;
  
    // Calculate power of all
    // prime factors of N
    for (int i = 2; i * i <= N; i++) {
  
        // Store the power of
        // prime factors
        int countPower = 0;
        while (N % i == 0) {
            countPower++;
            N /= i;
        }
  
        res = res * (2 * countPower
                     + 1);
    }
  
    if (N > 1) {
        res = res * (2 * 1 + 1);
    }
    return res;
}
  
// Driver Code
int main()
{
    int N = 36;
    cout << CtOrderedPairs(N);
}

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Java

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// Java program to implement
// the above approach
  
class GFG{
    
// Function to count the number of
// ordered pairs with given LCM
static int CtOrderedPairs(int N)
{
      
    // Stores count of
    // ordered pairs
    int res = 1;
  
    // Calculate power of all
    // prime factors of N
    for(int i = 2; i * i <= N; i++)
    {
          
        // Store the power of
        // prime factors
        int countPower = 0;
          
        while (N % i == 0
        {
            countPower++;
            N /= i;
        }
        res = res * (2 * countPower + 1);
    }
  
    if (N > 1)
    {
        res = res * (2 * 1 + 1);
    }
    return res;
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 36;
      
    System.out.println(CtOrderedPairs(N));
}
}
  
// This code is contributed by aimformohan

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Python3

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# Python3 program to implement
# the above approach
   
# Function to count the number of
# ordered pairs with given LCM
def CtOrderedPairs(N):
  
    # Stores count of
    # ordered pairs
    res = 1
   
    # Calculate power of all
    # prime factors of N
    i = 2 
    while(i * i <= N):
   
        # Store the power of
        # prime factors
        countPower = 0
        while (N % i == 0):
            countPower += 1
            N //= i
   
        res = res * (2 * countPower + 1)
        i += 1
      
    if (N > 1):
        res = res * (2 * 1 + 1)
      
    return res
  
# Driver Code
N = 36
  
print(CtOrderedPairs(N))
  
# This code is contributed by code_hunt

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C#

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// C# program to implement
// the above approach
using System;
   
class GFG{
     
// Function to count the number of
// ordered pairs with given LCM
static int CtOrderedPairs(int N)
{
      
    // Stores count of
    // ordered pairs
    int res = 1;
   
    // Calculate power of all
    // prime factors of N
    for(int i = 2; i * i <= N; i++)
    {
           
        // Store the power of
        // prime factors
        int countPower = 0;
           
        while (N % i == 0) 
        {
            countPower++;
            N /= i;
        }
        res = res * (2 * countPower + 1);
    }
   
    if (N > 1)
    {
        res = res * (2 * 1 + 1);
    }
    return res;
}
   
// Driver Code
public static void Main()
{
    int N = 36;
       
    Console.WriteLine(CtOrderedPairs(N));
}
}
  
// This code is contributed by code_hunt

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Output: 

25




 

Time Complexity: O(√N) 
Auxiliary Space: O(1)

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Improved By : aimformohan, code_hunt