Check if a given pair of Numbers are Betrothed numbers or not

Given two positive numbers N and M, the task is to check whether the given pairs of numbers (N, M) form a Betrothed Numbers or not.

Examples:

Input: N = 48, M = 75
Output: Yes
Explanation:
The proper divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24
Sum of proper divisors of 48 is 75(sum1)
The proper divisors of 75 are 1, 3, 5, 15, 25
Sum of proper divisors of 48 is 49(sum2)
Since sum2 = N + 1, therefore the given pairs form berothered numbers.

Input: N = 95, M = 55
Output: No
Explanation:
The proper divisors of 95 are 1, 5, 19
Sum of proper divisors of 48 is 25(sum1)
The proper divisors of 55 are 1, 5, 11
Sum of proper divisors of 48 is 17(sum2)
Since Neither sum2 is equals N + 1 nor sum1 is equals to M + 1, therefore the given pairs doesn’t form berothered numbers.

Approach:



  1. Find the sum of proper divisors of the given numbers N and M.
  2. If sum of proper divisors of N is equals to M + 1 or sum of proper divisors of M is equals to N + 1 then the given pairs form a Betrothed Numbers.
  3. Else it doen’t forms a pair of Betrothed Numbers.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to check whether N is
// Perfect Square or not
bool isPerfectSquare(int N)
{
  
    // Find sqrt
    double sr = sqrt(N);
  
    return (sr - floor(sr)) == 0;
}
  
// Function to check whether the given
// pairs of numbers is Betrothed Numbers
// or not
void BetrothedNumbers(int n, int m)
{
    int Sum1 = 1;
    int Sum2 = 1;
  
    // For finding the sum of all the
    // divisors of first number n
    for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) {
            Sum1 += i
                    + (isPerfectSquare(n)
                           ? 0
                           : n / i);
        }
    }
  
    // For finding the sum of all the
    // divisors of second number m
    for (int i = 2; i <= sqrt(m); i++) {
        if (m % i == 0) {
            Sum2 += i
                    + (isPerfectSquare(m)
                           ? 0
                           : m / i);
        }
    }
  
    if ((n + 1 == Sum2)
        && (m + 1 == Sum1)) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }
}
  
// Driver Code
int main()
{
    int N = 9504;
    int M = 20734;
  
    // Function Call
    BetrothedNumbers(N, M);
  
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
   
// Function to check whether N is
// Perfect Square or not
static boolean isPerfectSquare(int N)
{
   
    // Find sqrt
    double sr = Math.sqrt(N);
   
    return (sr - Math.floor(sr)) == 0;
}
   
// Function to check whether the given
// pairs of numbers is Betrothed Numbers
// or not
static void BetrothedNumbers(int n, int m)
{
    int Sum1 = 1;
    int Sum2 = 1;
   
    // For finding the sum of all the
    // divisors of first number n
    for (int i = 2; i <= Math.sqrt(n); i++) {
        if (n % i == 0) {
            Sum1 += i
                    + (isPerfectSquare(n)
                           ? 0
                           : n / i);
        }
    }
   
    // For finding the sum of all the
    // divisors of second number m
    for (int i = 2; i <= Math.sqrt(m); i++) {
        if (m % i == 0) {
            Sum2 += i
                    + (isPerfectSquare(m)
                           ? 0
                           : m / i);
        }
    }
   
    if ((n + 1 == Sum2)
        && (m + 1 == Sum1)) {
        System.out.print("YES" +"\n");
    }
    else {
        System.out.print("NO" +"\n");
    }
}
   
// Driver Code
public static void main(String[] args)
{
    int N = 9504;
    int M = 20734;
   
    // Function Call
    BetrothedNumbers(N, M);
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program for the above approach
from math import sqrt,floor
  
# Function to check whether N is
# Perfect Square or not
def isPerfectSquare(N):
    # Find sqrt
    sr = sqrt(N)
  
    return (sr - floor(sr)) == 0
  
# Function to check whether the given
# pairs of numbers is Betrothed Numbers
# or not
def BetrothedNumbers(n,m):
    Sum1 = 1
    Sum2 = 1
  
    # For finding the sum of all the
    # divisors of first number n
    for i in range(2,int(sqrt(n))+1,1):
        if (n % i == 0):
            if (isPerfectSquare(n)):
                Sum1 += i
            else:
                Sum1 += i + n/i
  
    # For finding the sum of all the
    # divisors of second number m
    for i in range(2,int(sqrt(m))+1,1):
        if (m % i == 0):
            if (isPerfectSquare(m)):
                Sum2 += i
            else:
                Sum2 += i + (m / i)
  
    if ((n + 1 == Sum2) and (m + 1 == Sum1)):
        print("YES")    
    else:
        print("NO")
  
# Driver Code
if __name__ == '__main__':
    N = 9504
    M = 20734
  
    # Function Call
    BetrothedNumbers(N, M)
  
# This code is contributed by Surendra_Gangwar

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C#

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// C# program for the above approach
using System;
  
class GFG{
  
// Function to check whether N is
// perfect square or not
static bool isPerfectSquare(int N)
{
  
    // Find sqrt
    double sr = Math.Sqrt(N);
  
    return (sr - Math.Floor(sr)) == 0;
}
  
// Function to check whether the given
// pairs of numbers is Betrothed numbers
// or not
static void BetrothedNumbers(int n, int m)
{
    int Sum1 = 1;
    int Sum2 = 1;
  
    // For finding the sum of all the
    // divisors of first number n
    for(int i = 2; i <= Math.Sqrt(n); i++)
    {
       if (n % i == 0)
       {
           Sum1 += i + (isPerfectSquare(n) ?
                                 0 : n / i);
       }
    }
  
    // For finding the sum of all the
    // divisors of second number m
    for(int i = 2; i <= Math.Sqrt(m); i++) 
    {
       if (m % i == 0)
       {
           Sum2 += i + (isPerfectSquare(m) ?
                                 0 : m / i);
       }
    }
  
    if ((n + 1 == Sum2) && (m + 1 == Sum1))
    {
        Console.Write("YES" + "\n");
    }
    else
    {
        Console.Write("NO" + "\n");
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    int N = 9504;
    int M = 20734;
  
    // Function Call
    BetrothedNumbers(N, M);
}
}
  
// This code is contributed by Rajput-Ji

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Output:

NO

Time Complexity: O(√N + √M)

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