Find unique pairs such that each element is less than or equal to N

Given an integer N, find and show the number of pairs which satisfies the following conditions:

  • Square of distance between those two numbers is equal to the LCM of those two numbers.
  • The GCD of those two numbers is equal to the product of two consecutive integers.
  • Both numbers in the pair should be less than or equal to N.

NOTE: Only those pairs should be displayed which follows both the above conditions simultaneously and those numbers must be less than or equal to N.

Examples:

Input: 10
Output: No. of pairs = 1
        Pair no. 1 --> (2, 4)

Input: 500
Output: No. of pairs = 7
        Pair no. 1 --> (2, 4)
        Pair no. 2 --> (12, 18)
        Pair no. 3 --> (36, 48)
        Pair no. 4 --> (80, 100)
        Pair no. 5 --> (150, 180)
        Pair no. 6 --> (252, 294)
        Pair no. 7 --> (392, 448)

Explanation:

The tables shown below will give a clear view of what is to be found :

Above tables show GCD formed by the product of two consecutive numbers and its corresponding multiples in which UNIQUE PAIR exists corresponding to each value. Green entries in each row form a unique pair for corresponding GCD.

Note: In the above tables,

  1. For 1st entry, GCD=2, 1st and the 2nd multiple of 2 form the Unique Pair, (2, 4)
  2. Similarly, for the 2nd entry, GCD=6, 2nd and the 3rd multiple of 6 form the Unique Pair, (12, 18)
  3. Similarly, moving on, for Zth entry, i.e for GCD = Z*(Z+1), it is clear that the unique pair will comprise of Zth and (Z+1)th multiple of GCD = Z*(Z+1). Now, Zth multiple of GCD is Z * (Z*(Z+1)) and (Z+1)th multiple of GCD will be (Z + 1) * (Z*(Z+1)).
  4. And as the limit is N, so the second number in the unique pair must be less than or equal to the N. So, (Z + 1) * (Z*(Z+1)) <= N. Simplifying it further, the desired relation is derived Z3 + (2*Z2) + Z <=N

This forms a pattern and from the mathematical calculation, it is derived that for a given N, the total number of such unique pairs (say, Z) will follow a mathematical relation shown below:

Z3 + (2*Z2) + Z <= N

Below is the required implementation:

C

filter_none

edit
close

play_arrow

link
brightness_4
code

// C program for finding the required pairs
#include <stdio.h>
#include <stdlib.h>
  
// Finding the number of unique pairs
int No_Of_Pairs(int N)
{
    int i = 1;
  
    // Using the derived formula
    while ((i * i * i) + (2 * i * i) + i <= N)
        i++;
  
    return (i - 1);
}
  
// Printing the unique pairs
void print_pairs(int pairs)
{
    int i = 1, mul;
    for (i = 1; i <= pairs; i++) {
        mul = i * (i + 1);
        printf("Pair no. %d --> (%d, %d)\n",
               i, (mul * i), mul * (i + 1));
    }
}
  
// Driver program to test above functions
int main()
{
    int N = 500, pairs, mul, i = 1;
    pairs = No_Of_Pairs(N);
  
    printf("No. of pairs = %d \n", pairs);
    print_pairs(pairs);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for finding
// the required pairs
import java.io.*;
  
class GFG 
{
      
    // Finding the number
    // of unique pairs
    static int No_Of_Pairs(int N)
    {
        int i = 1;
      
        // Using the derived formula
        while ((i * i * i) + 
               (2 * i * i) + i <= N)
            i++;
      
        return (i - 1);
    }
      
    // Printing the unique pairs
    static void print_pairs(int pairs)
    {
        int i = 1, mul;
        for (i = 1; i <= pairs; i++)
        {
            mul = i * (i + 1);
            System.out.println("Pair no. " + i + " --> ("
                                         (mul * i) + ", "
                                      mul * (i + 1) + ")"); 
        }
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int N = 500, pairs, mul, i = 1;
        pairs = No_Of_Pairs(N);
      
        System.out.println("No. of pairs = " + pairs);
        print_pairs(pairs);
    }
}
  
// This code is contributed by Mahadev.

chevron_right


Python3

# Python3 program for finding the required pairs

# Finding the number of unique pairs
def No_Of_Pairs(N):

i = 1;

# Using the derived formula
while ((i * i * i) + (2 * i * i) + i <= N): i += 1; return (i - 1); # Printing the unique pairs def print_pairs(pairs): i = 1; mul = 0; for i in range(1, pairs + 1): mul = i * (i + 1); print("Pair no." , i, " --> (“, (mul * i),
“, “, mul * (i + 1), “)”);

# Driver Code
N = 500;
i = 1;
pairs = No_Of_Pairs(N);

print(“No. of pairs = “, pairs);
print_pairs(pairs);

# This code is contributed
# by mits

C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for finding
// the required pairs
using System;
  
class GFG 
{
      
// Finding the number
// of unique pairs
static int No_Of_Pairs(int N)
{
    int i = 1;
  
    // Using the derived formula
    while ((i * i * i) + 
           (2 * i * i) + i <= N)
        i++;
  
    return (i - 1);
}
  
// Printing the unique pairs
static void print_pairs(int pairs)
{
    int i = 1, mul;
    for (i = 1; i <= pairs; i++)
    {
        mul = i * (i + 1);
        Console.WriteLine("Pair no. " + i + " --> ("
                                    (mul * i) + ", "
                                 mul * (i + 1) + ")"); 
    }
}
  
// Driver code
static void Main()
{
    int N = 500, pairs;
    pairs = No_Of_Pairs(N);
  
    Console.WriteLine("No. of pairs = "
                                  pairs);
    print_pairs(pairs);
}
}
  
// This code is contributed by mits

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program for finding 
// the required pairs
  
// Finding the number 
// of unique pairs
function No_Of_Pairs($N)
{
    $i = 1;
  
    // Using the 
    // derived formula
    while (($i * $i * $i) + 
            (2 * $i * $i) + 
                $i <= $N)
        $i++;
  
    return ($i - 1);
}
  
// Printing the unique pairs
function print_pairs($pairs)
{
    $i = 1; $mul;
    for ($i = 1; 
         $i <= $pairs; $i++)
    {
        $mul = $i * ($i + 1);
        echo "Pair no."
              $i, " --> (" ,
             ($mul * $i), ", ",
              $mul * ($i + 1),") \n";
    }
}
  
// Driver Code
$N = 500; $pairs
$mul; $i = 1;
$pairs = No_Of_Pairs($N);
  
echo "No. of pairs = "
        $pairs , " \n";
print_pairs($pairs);
  
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

chevron_right


Output:

No. of pairs = 7 
Pair no. 1 --> (2, 4)
Pair no. 2 --> (12, 18)
Pair no. 3 --> (36, 48)
Pair no. 4 --> (80, 100)
Pair no. 5 --> (150, 180)
Pair no. 6 --> (252, 294)
Pair no. 7 --> (392, 448)


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.