Primitive Abundant Number

A number N is said to be Primitive Abundant Number if N is an Abundant number and all it’s proper divisors are Deficient Numbers. 
The first few Primitive Abundant Numbers are:

20, 70, 88, 104, 272, 304……… 

Check if N is a Primitive Abundant Number

Given a number N, the task is to find if this number is Primitive Abundant Number or not. 
Examples:

Input: N = 20 
Output: YES 
Explanation:
Sum of 20’s proper divisors is – 1 + 2 + 4 + 5 + 10 = 22 > 20, 
So, 20 is an abundant number. 
The proper divisors of 1, 2, 4, 5 and 10 are0, 1, 3, 1 and 8 respectively, 
Each of these numbers is a deficient number 
Therefore, 20 is a primitive abundant number.
Input: N = 17 
Output: No 

Approach:

1. Check if the number is an Abundant number or not, i.e, sum of all the proper divisors of the number denoted by sum(N) is greater than the value of the number N
2. If the number is not abundant then return false else do the following
3. Check if all proper divisors of N are Deficient Numbers or not, i.e, sum of all the divisors of the number denoted by divisorsSum(n) is less than twice the value of the number N.
4. If both above conditions are true print “Yes” else print “No.

Below is the implementation of the above approach:

C++

// C++ implementation of the above
// approach

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

// Function to sum of divisors
int getSum(int n)
{
    int sum = 0;

    // Note that this loop
    // runs till square root of N
    for (int i = 1; i <= sqrt(n); i++) {

        if (n % i == 0) {

            // If divisors are equal,
            // take only one of them
            if (n / i == i)
                sum = sum + i;

            else // Otherwise take both
            {
                sum = sum + i;
                sum = sum + (n / i);
            }
        }
    }

    return sum;
}

// Function to check Abundant Number
bool checkAbundant(int n)
{
    // Return true if sum
    // of divisors is greater
    // than N.
    return (getSum(n) - n > n);
}

// Function to check Deficient Number
bool isDeficient(int n)
{
    // Check if sum(n) < 2 * n
    return (getSum(n) < (2 * n));
}

// Function to check all proper divisors
// of N is deficient number or not
bool checkPrimitiveAbundant(int num)
{
    // if number itself is not abundant
    // return false
    if (!checkAbundant(num)) {
        return false;
    }

    // find all divisors which divides 'num'
    for (int i = 2; i <= sqrt(num); i++) {

        // if 'i' is divisor of 'num'
        if (num % i == 0 && i != num) {

            // if both divisors are same then add
            // it only once else add both
            if (i * i == num) {
                if (!isDeficient(i)) {
                    return false;
                }
            }
            else if (!isDeficient(i) || !isDeficient(num / i)) {
                return false;
            }
        }
    }

    return true;
}

// Driver Code
int main()
{

    int n = 20;
    if (checkPrimitiveAbundant(n)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java implementation of the above
// approach
class GFG{

// Function to sum of divisors
static int getSum(int n)
{
    int sum = 0;

    // Note that this loop runs
    // till square root of N
    for(int i = 1; i <= Math.sqrt(n); i++)
    {
       if (n % i == 0)
       {

           // If divisors are equal,
           // take only one of them
           if (n / i == i)
               sum = sum + i;

           // Otherwise take both
           else
           {
               sum = sum + i;
               sum = sum + (n / i);
           }
       }
    }
    return sum;
}

// Function to check Abundant Number
static boolean checkAbundant(int n)
{

    // Return true if sum
    // of divisors is greater
    // than N.
    return (getSum(n) - n > n);
}

// Function to check Deficient Number
static boolean isDeficient(int n)
{

    // Check if sum(n) < 2 * n
    return (getSum(n) < (2 * n));
}

// Function to check all proper divisors
// of N is deficient number or not
static boolean checkPrimitiveAbundant(int num)
{

    // If number itself is not abundant
    // return false
    if (!checkAbundant(num))
    {
        return false;
    }

    // Find all divisors which divides 'num'
    for(int i = 2; i <= Math.sqrt(num); i++)
    {

       // if 'i' is divisor of 'num'
       if (num % i == 0 && i != num)
       {

           // if both divisors are same then
           // add it only once else add both
           if (i * i == num)
           {
               if (!isDeficient(i))
               {
                   return false;
               }
           }
           else if (!isDeficient(i) ||
                    !isDeficient(num / i))
           {
               return false;
           }
       }
    }
    return true;
}

// Driver Code
public static void main(String[] args)
{
    int n = 20;

    if (checkPrimitiveAbundant(n))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}

// This code is contributed by Ritik Bansal

Python3

# Python3 implementation of the above
# approach
import math

# Function to sum of divisors
def getSum(n):
    sum = 0

    # Note that this loop
    # runs till square root of N
    for i in range(1, int(math.sqrt(n) + 1)):
        if (n % i == 0):

            # If divisors are equal,
            # take only one of them
            if (n // i == i):
                sum = sum + i
            else:

                # Otherwise take both
                sum = sum + i
                sum = sum + (n // i)
    return sum

# Function to check Abundant Number
def checkAbundant(n):

    # Return True if sum
    # of divisors is greater
    # than N.
    if (getSum(n) - n > n):
        return True
    return False

# Function to check Deficient Number
def isDeficient(n):

    # Check if sum(n) < 2 * n
    if (getSum(n) < (2 * n)):
        return True
    return False

# Function to check all proper divisors
# of N is deficient number or not
def checkPrimitiveAbundant(num):

    # if number itself is not abundant
    # return False
    if not checkAbundant(num):
        return False

    # find all divisors which divides 'num'
    for i in range(2, int(math.sqrt(num) + 1)):

        # if 'i' is divisor of 'num'
        if (num % i == 0 and i != num):
            # if both divisors are same then add
            # it only once else add both
            if (i * i == num):
                if (not isDeficient(i)):
                    return False
            elif (not isDeficient(i) or
                  not isDeficient(num // i)):
                return False
    return True

# Driver Code
n = 20
if (checkPrimitiveAbundant(n)):
    print("Yes")
else:
    print("No")

# This code is contributed by shubhamsingh10

C#

// C# implementation of the above
// approach
using System;
class GFG{

// Function to sum of divisors
static int getSum(int n)
{
    int sum = 0;

    // Note that this loop runs
    // till square root of N
    for(int i = 1; i <= Math.Sqrt(n); i++)
    {
       if (n % i == 0)
       {

           // If divisors are equal,
           // take only one of them
           if (n / i == i)
               sum = sum + i;

           // Otherwise take both
           else
           {
               sum = sum + i;
               sum = sum + (n / i);
           }
       }
    }
    return sum;
}

// Function to check Abundant Number
static bool checkAbundant(int n)
{

    // Return true if sum
    // of divisors is greater
    // than N.
    return (getSum(n) - n > n);
}

// Function to check Deficient Number
static bool isDeficient(int n)
{

    // Check if sum(n) < 2 * n
    return (getSum(n) < (2 * n));
}

// Function to check all proper divisors
// of N is deficient number or not
static bool checkPrimitiveAbundant(int num)
{

    // If number itself is not abundant
    // return false
    if (!checkAbundant(num))
    {
        return false;
    }

    // Find all divisors which divides 'num'
    for(int i = 2; i <= Math.Sqrt(num); i++)
    {

       // If 'i' is divisor of 'num'
       if (num % i == 0 && i != num)
       {

           // If both divisors are same then
           // add it only once else add both
           if (i * i == num)
           {
               if (!isDeficient(i))
               {
                   return false;
               }
           }
           else if (!isDeficient(i) ||
                    !isDeficient(num / i))
           {
               return false;
           }
       }
    }
    return true;
}

// Driver Code
public static void Main()
{
    int n = 20;

    if (checkPrimitiveAbundant(n))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}

// This code is contributed by Code_Mech

Javascript

<script>
// Javascript implementation of the above
// approach


    // Function to sum of divisors
    function getSum( n) {
        let sum = 0;

        // Note that this loop runs
        // till square root of N
        for ( let i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {

                // If divisors are equal,
                // take only one of them
                if (n / i == i)
                    sum = sum + i;

                // Otherwise take both
                else {
                    sum = sum + i;
                    sum = sum + (n / i);
                }
            }
        }
        return sum;
    }

    // Function to check Abundant Number
    function checkAbundant( n) {

        // Return true if sum
        // of divisors is greater
        // than N.
        return (getSum(n) - n > n);
    }

    // Function to check Deficient Number
    function isDeficient( n) {

        // Check if sum(n) < 2 * n
        return (getSum(n) < (2 * n));
    }

    // Function to check all proper divisors
    // of N is deficient number or not
    function checkPrimitiveAbundant( num) {

        // If number itself is not abundant
        // return false
        if (!checkAbundant(num)) {
            return false;
        }

        // Find all divisors which divides 'num'
        for ( let i = 2; i <= Math.sqrt(num); i++) {

            // if 'i' is divisor of 'num'
            if (num % i == 0 && i != num) {

                // if both divisors are same then
                // add it only once else add both
                if (i * i == num) {
                    if (!isDeficient(i)) {
                        return false;
                    }
                } else if (!isDeficient(i) || !isDeficient(num / i)) {
                    return false;
                }
            }
        }
        return true;
    }

    // Driver Code

        let n = 20;

        if (checkPrimitiveAbundant(n)) {
            document.write("Yes");
        } else {
            document.write("No");
        }

// This code contributed by aashish1995
</script>

Output:

Yes

Time Complexity: O(N^1/2)

Auxiliary Space: O(1)

References: https://en.wikipedia.org/wiki/Primitive_abundant_number