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# Count ordered pairs of numbers with a given LCM

• Difficulty Level : Medium
• Last Updated : 10 May, 2021

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++

 `// C++ program to implement``// the above approach` `#include ``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

 ``
Output:
`25`

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

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