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: 9 
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

Brute Force Approach:

A brute force approach to solve this problem would be to consider all possible pairs of integers (a,b) such that 1 <= a,b <= N, and then check if their LCM is equal to N. If it is, then count it as a valid pair.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// Function to count the number of
// ordered pairs with given LCM
int CtOrderedPairs(int N)
{
    int count = 1;
    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= N; j++) {
            if (i != j && i * j / __gcd(i, j) == N) {

                // Increment the count
                count++;
            }
        }
    }
    return count;
}

// Driver Code
int main()
{
    int N = 6;
    cout << CtOrderedPairs(N);
    return 0;
}

Python3

# Import the gcd function from the math module
from math import gcd

# Function to count the number of ordered pairs with given LCM


def count_ordered_pairs(N):
    count = 1
    for i in range(1, N+1):
        for j in range(1, N+1):
            if i != j and i*j // gcd(i, j) == N:
                count += 1
    return count


# Driver code
N = 6
print(count_ordered_pairs(N))

C#

using System;

public class Program {

    // Function to count the number of
    // ordered pairs with given LCM
    public static int CtOrderedPairs(int N)
    {
        int count = 1;
        for (int i = 1; i <= N; i++) {
            for (int j = 1; j <= N; j++) {
                if (i != j && i * j / GCD(i, j) == N) {
                    count++;
                }
            }
        }
        return count; // To avoid counting same pairs twice
    }

    // Function to calculate GCD of two numbers
    public static int GCD(int a, int b)
    {
        if (b == 0)
            return a;
        return GCD(b, a % b);
    }

    // Driver Code
    public static void Main()
    {
        int N = 6;
        Console.WriteLine(CtOrderedPairs(N));
    }
}

Java

import java.util.*;

public class Main {
    // Function to count the number of
    // ordered pairs with given LCM
    static int ctOrderedPairs(int N)
    {
        int count = 1;
        for (int i = 1; i <= N; i++) {
            for (int j = 1; j <= N; j++) {
                if (i != j && i * j / gcd(i, j) == N) {
                    count++;
                }
            }
        }
        return count; // To avoid counting same pairs twice
    }

    // Function to calculate GCD using Euclidean algorithm
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        System.out.println(ctOrderedPairs(N));
    }
}

Javascript

// Function to count the number of ordered pairs with given LCM
function count_ordered_pairs(N) {
let count = 1;
for (let i = 1; i <= N; i++) {
for (let j = 1; j <= N; j++) {
if (i !== j && (i*j) / gcd(i, j) === N) {
count += 1;
}
}
}
return count;
}

// Function to calculate GCD using Euclidean algorithm
function gcd(a, b) {
if (b === 0) {
return a;
}
return gcd(b, a % b);
}


// Driver code
const N = 6;
console.log(count_ordered_pairs(N));

Output

9

Time Complexity: O(N^2)

Auxiliary Space: O(1)

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

Consider an ordered pair(X, Y). 
X = P₁^a1 * P₂^a2 * P₃^a3 *…..* Pn^an
Y = P₁^b1 * P₂^b2 * P₃^b3 *…..* Pn^bn
Here, P₁, P₂, ….., P_n are prime factors of X and Y. 
LCM(X, Y) = P₁^{max(a1, b1)}  * P2^{max(a2, b2)} *……….*Pn^{max(an, bn)}
Therefore, LCM(X, Y) = N = P₁^m₁ * P₂^m₂ * P₃^m₃ *…..* Pn^m_n

Therefore, total number of ordered pairs (X, Y) 
= [{(m₁ + 1)² – m₁²} * {(m₂ + 1)² – m₂²} * ……* {(m_n + 1)² – m_n²} ]
= (2*m₁+1) * (2*m₂+1) * (2*m₃+1) * ……..* (2*m_n+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++

// 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);
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to count the number of
// ordered pairs with given LCM
function CtOrderedPairs(N)
{

    // Stores count of
    // ordered pairs
    let res = 1;

    // Calculate power of all
    // prime factors of N
    for(let i = 2; i * i <= N; i++)
    {

        // Store the power of
        // prime factors
        let 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

    let N = 36;

    document.write(CtOrderedPairs(N));

</script>

Output

25

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