# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 ` `    ``// retuen 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 ` `    ``// retuen 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 ` `    ``# retuen 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 ` `    ``// retuen 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 `

Output:

```Yes
```

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