Find the sum of all Betrothed numbers up to N

Last Updated : 16 Nov, 2020

Given a number N, the task is to find the sum of the all Betrothed numbers up to N.

Let, A and B are a Betrothed number pair, then the sum of the proper divisors of A is equal to B+1 and the sum of the proper divisors of B is equal to A+1.

Examples:

Input: 78
Output: 123
Explanation: 48 and 75 is the only pair of Betrothed Numbers upto 78.

Input: 5
Output: 0
Explanation: There is no pair of Betrothed Numbers up to 50.

Approach:

Initialise the array to store all Betrothed numbers upto to N
Find the Betrothed numbers using sum of all its proper divisors and store the pair in the array.
Then find the sum all the numbers less than and equal to N.
The resultant sum is the required value.

Below code is the implementation of the above approach:

C++

// C++ program to find the sum of the   
// all betrothed numbers up to N 
#include<bits/stdc++.h>
using namespace std;
 
// Function to find the sum of 
// the all betrothed numbers 
int Betrothed_Sum(int n)
{
    // To store the betrothed 
    // numbers 
    vector<int > Set;
     
    for(int number_1 = 1; number_1 < n; 
                            number_1++)
    {
         
       // Calculate sum of 
       // number_1's divisors 
       // 1 is always a divisor
       int sum_divisor_1 = 1;
        
       // i = 2 because we don't 
       // want to include 
       // 1 as a divisor. 
       int i = 2;
        
       while (i * i <= number_1)
       {
           if (number_1 % i == 0)
           {
               sum_divisor_1 = sum_divisor_1 + i;
                
               if (i * i != number_1)
                   sum_divisor_1 += number_1 / i; 
           }
           i ++;
       }
       if (sum_divisor_1 > number_1)
       {
           int number_2 = sum_divisor_1 - 1;
           int sum_divisor_2 = 1;
           int j = 2;
            
           while (j * j <= number_2) 
           {
               if (number_2 % j == 0) 
               {
                   sum_divisor_2 += j;
                   if (j * j != number_2)
                       sum_divisor_2 += number_2 / j;
               }
               j = j + 1;
           }
           if (sum_divisor_2 == number_1 + 1 and 
               number_1 <= n && number_2 <= n)
           {
               Set.push_back(number_1);
               Set.push_back(number_2);
           }
       }
    }
     
    // Sum all betrothed 
    // numbers up to N 
    int Summ = 0;
    for(auto i : Set)
    {
       if(i <= n) 
          Summ += i;
    }
    return Summ;
}
 
// Driver code 
int main()
{
    int n = 78;
     
    cout << Betrothed_Sum(n);
    return 0;
}
 
// This code is contributed by ishayadav181

Java

// Java program to find the sum
// of the all betrothed numbers
// up to N  
import java.util.*;
 
class GFG{
     
// Function to find the sum of  
// the all betrothed numbers  
public static int Betrothed_Sum(int n) 
{ 
     
    // To store the betrothed  
    // numbers  
    Vector<Integer> Set = new Vector<Integer>(); 
       
    for(int number_1 = 1; 
            number_1 < n;  
            number_1++) 
    { 
         
       // Calculate sum of  
       // number_1's divisors  
       // 1 is always a divisor 
       int sum_divisor_1 = 1; 
          
       // i = 2 because we don't  
       // want to include  
       // 1 as a divisor.  
       int i = 2; 
          
       while (i * i <= number_1) 
       { 
           if (number_1 % i == 0) 
           { 
               sum_divisor_1 = sum_divisor_1 + i; 
                  
               if (i * i != number_1) 
                   sum_divisor_1 += number_1 / i;  
           } 
           i ++; 
       } 
        
       if (sum_divisor_1 > number_1) 
       { 
           int number_2 = sum_divisor_1 - 1; 
           int sum_divisor_2 = 1; 
           int j = 2; 
              
           while (j * j <= number_2)  
           { 
               if (number_2 % j == 0)  
               { 
                   sum_divisor_2 += j; 
                    
                   if (j * j != number_2) 
                       sum_divisor_2 += number_2 / j; 
               } 
               j = j + 1; 
           } 
           if (sum_divisor_2 == number_1 + 1 &&  
               number_1 <= n && number_2 <= n) 
           { 
               Set.add(number_1); 
               Set.add(number_2); 
           } 
       } 
    } 
       
    // Sum all betrothed  
    // numbers up to N  
    int Summ = 0; 
    for(int i = 0; i < Set.size(); i++) 
    { 
       if (Set.get(i) <= n)  
          Summ += Set.get(i); 
    } 
    return Summ; 
} 
 
// Driver code
public static void main(String[] args) 
{
    int n = 78; 
     
    System.out.println(Betrothed_Sum(n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find the 
# Sum of the all Betrothed 
# numbers up to N
 
import math
 
# Function to find the Sum of
# the all Betrothed numbers
def Betrothed_Sum(n):
     
    # To store the Betrothed 
    # numbers
    Set = []
 
    for number_1 in range(1, n):
 
        # Calculate sum of 
        # number_1's divisors
 
        # 1 is always a divisor
        sum_divisor_1 = 1
 
        # i = 2 because we don't
        # want to include
        # 1 as a divisor.
        i = 2
 
        while i * i <= number_1:
            if (number_1 % i == 0):
                sum_divisor_1 = sum_divisor_1 + i
 
                if (i * i != number_1):
                    sum_divisor_1 += number_1 // i
            i = i + 1
 
        if (sum_divisor_1 > number_1):
 
            number_2 = sum_divisor_1 - 1
            sum_divisor_2 = 1
            j = 2
            while j * j <= number_2:
                if (number_2 % j == 0):
                    sum_divisor_2 += j
                    if (j * j != number_2):
                        sum_divisor_2 += number_2 // j
                j = j + 1
 
            if (sum_divisor_2 == number_1 + 1
                and number_1<= n and number_2<= n):
                Set.append(number_1)
                Set.append(number_2)
 
    # Sum all Betrothed
    # numbers up to N
    Summ = 0
    for i in Set:
        if i <= n:
            Summ += i
    return Summ
 
 
# Driver Code
n = 78
print(Betrothed_Sum(n))

C#

// C# program to find the sum
// of the all betrothed numbers
// up to N  
using System;
using System.Collections;
 
class GFG{
     
// Function to find the sum of  
// the all betrothed numbers  
public static int Betrothed_Sum(int n) 
{ 
     
    // To store the betrothed  
    // numbers  
    ArrayList set = new ArrayList(); 
        
    for(int number_1 = 1; 
            number_1 < n;  
            number_1++) 
    { 
         
        // Calculate sum of  
        // number_1's divisors  
        // 1 is always a divisor 
        int sum_divisor_1 = 1; 
         
        // i = 2 because we don't  
        // want to include  
        // 1 as a divisor.  
        int i = 2; 
         
        while (i * i <= number_1) 
        { 
            if (number_1 % i == 0) 
            { 
                sum_divisor_1 = sum_divisor_1 + i; 
                 
                if (i * i != number_1) 
                    sum_divisor_1 += number_1 / i;  
            } 
            i ++; 
        } 
         
        if (sum_divisor_1 > number_1) 
        { 
            int number_2 = sum_divisor_1 - 1; 
            int sum_divisor_2 = 1; 
            int j = 2; 
             
            while (j * j <= number_2)  
            { 
                if (number_2 % j == 0)  
                { 
                    sum_divisor_2 += j; 
                     
                    if (j * j != number_2) 
                        sum_divisor_2 += number_2 / j; 
                } 
                j = j + 1; 
            } 
             
            if (sum_divisor_2 == number_1 + 1 &&  
                number_1 <= n && number_2 <= n) 
            { 
                set.Add(number_1); 
                set.Add(number_2); 
            } 
        } 
    } 
        
    // Sum all betrothed  
    // numbers up to N  
    int Summ = 0; 
    for(int i = 0; i < set.Count; i++) 
    { 
        if ((int)set[i] <= n)  
            Summ += (int)set[i]; 
    } 
    return Summ; 
}
 
// Driver code
static public void Main()
{
    int n = 78; 
      
    Console.WriteLine(Betrothed_Sum(n));
}
}
 
// This code is contributed by offbeat

Output:

Time Complexity: O(N * sqrt(N))

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