# Determine whether a given number is a Hyperperfect Number

Given a number, determine whether it is a valid Hyperperfect Number.
A number n is called k-hyperperfect if: n = 1 + k ∑idi where all di are the proper divisors of n.
Taking k = 1 will give us perfect numbers.

The first few k-hyperperfect numbers are 6, 21, 28, 301, 325, 496, 697, … with the corresponding values of k being 1, 2, 1, 6, 3, 1, 12, …

Examples:

```Input :  N = 36, K = 1
Output :  34 is not 1-HyperPerfect
Explanation:
The Divisors of 36 are 2, 3, 4, 6, 9, 12, 18
the sum of the divisors is 54.
For N = 36 to be 1-Hyperperfect, it would
require 36 = 1 + 1(54), which we see, is
invalid

Input :  N = 325, K = 3
Output :  325 is 3-HyperPerfect
Explanation:
We can use the first condition to evaluate this
as K is odd and > 1 so here p = (3*k+1)/2 = 5,
q = (3*k+4) = 13 p and q are both prime, so we
compute p^2 * q = 5 ^ 2 * 13 = 325
Hence N is a valid HyperPerfect number
```

## C++

 `// C++ 4.3.2 program to check whether a ` `// given number is  k-hyperperfect ` `#include ` `using` `namespace` `std; ` ` `  `// function to find the sum of all ` `// proper divisors (excluding 1 and N) ` `int` `divisorSum(``int` `N, ``int` `K) ` `{ ` `    ``int` `sum = 0; ` ` `  `    ``// Iterate only until sqrt N as we are ` `    ``// going to generate pairs to produce ` `    ``// divisors ` `    ``for` `(``int` `i = 2 ; i <= ``ceil``(``sqrt``(N)) ; i++) ` ` `  `        ``// As divisors occur in pairs, we can ` `        ``// take the values i and N/i as long ` `        ``// as i divides N ` `        ``if` `(N % i == 0) ` `            ``sum += ( i + N/i ); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Function to check whether the given number ` `// is prime ` `bool` `isPrime(``int` `n) ` `{ ` `    ``//base and corner cases ` `    ``if` `(n == 1 || n == 0) ` `        ``return` `false``; ` ` `  `    ``if` `(n <= 3) ` `        ``return` `true``; ` ` `  `    ``// Since integers can be represented as ` `    ``// some 6*k + y where y >= 0, we can eliminate ` `    ``// all integers that can be expressed in this ` `    ``// form ` `    ``if` `(n % 2 == 0 || n % 3 == 0) ` `        ``return` `false``; ` ` `  `    ``// start from 5 as this is the next prime number ` `    ``for` `(``int` `i=5; i*i<=n; i=i+6) ` `        ``if` `(n % i == 0 || n % ( i + 2 ) == 0) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Returns true if N is a K-Hyperperfect number ` `// Else returns false. ` `bool` `isHyperPerfect(``int` `N, ``int` `K) ` `{ ` `    ``int` `sum = divisorSum(N, K); ` ` `  `    ``// Condition from the definition of hyperperfect ` `    ``if` `((1 + K * (sum)) == N) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver function to test for hyperperfect numbers ` `int` `main() ` `{ ` `    ``int` `N1 = 1570153, K1 = 12; ` `    ``int` `N2 = 321, K2 = 3; ` ` `  `    ``// First two statements test against the condition ` `    ``// N = 1 + K*(sum(proper divisors)) ` `    ``if` `(isHyperPerfect(N1, K1)) ` `        ``cout << N1 << ``" is "` `<< K1  ` `             ``<<``"-HyperPerfect"` `<< ``"\n"``; ` `    ``else` `        ``cout << N1 << ``" is not "` `<< K1  ` `             ``<<``"-HyperPerfect"` `<< ``"\n"``; ` ` `  `    ``if` `(isHyperPerfect(N2, K2)) ` `        ``cout << N2 << ``" is "` `<< K2  ` `             ``<<``"-HyperPerfect"` `<< ``"\n"``; ` `    ``else` `        ``cout << N2 << ``" is not "` `<< K2  ` `             ``<<``"-HyperPerfect"` `<< ``"\n"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to check  ` `// whether a given number ` `// is k-hyperperfect ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// function to find the  ` `// sum of all proper  ` `// divisors (excluding  ` `// 1 and N) ` `static` `int` `divisorSum(``int` `N,  ` `                      ``int` `K) ` `{ ` `    ``int` `sum = ``0``; ` ` `  `    ``// Iterate only until  ` `    ``// sqrt N as we are  ` `    ``// going to generate  ` `    ``// pairs to produce ` `    ``// divisors ` `    ``for` `(``int` `i = ``2` `; ` `             ``i <= Math.ceil(Math.sqrt(N));  ` `             ``i++) ` ` `  `        ``// As divisors occur in  ` `        ``// pairs, we can take  ` `        ``// the values i and N/i  ` `        ``// as long as i divides N ` `        ``if` `(N % i == ``0``) ` `            ``sum += (i + N / i); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Function to check  ` `// whether the given ` `// number is prime ` `static` `boolean` `isPrime(``int` `n) ` `{ ` `    ``// base and corner cases ` `    ``if` `(n == ``1` `|| n == ``0``) ` `        ``return` `false``; ` ` `  `    ``if` `(n <= ``3``) ` `        ``return` `true``; ` ` `  `    ``// Since integers can be  ` `    ``// represented as some  ` `    ``// 6*k + y where y >= 0,  ` `    ``// we can eliminate all  ` `    ``// integers that can be  ` `    ``// expressed in this form ` `    ``if` `(n % ``2` `== ``0` `||  ` `        ``n % ``3` `== ``0``) ` `        ``return` `false``; ` ` `  `    ``// start from 5 as this  ` `    ``// is the next prime number ` `    ``for` `(``int` `i = ``5``;  ` `             ``i * i <= n; i = i + ``6``) ` `        ``if` `(n % i == ``0` `||  ` `            ``n % ( i + ``2` `) == ``0``) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Returns true if N is  ` `// a K-Hyperperfect number ` `// Else returns false. ` `static` `boolean` `isHyperPerfect(``int` `N, ` `                              ``int` `K) ` `{ ` `    ``int` `sum = divisorSum(N, K); ` ` `  `    ``// Condition from the  ` `    ``// definition of hyperperfect ` `    ``if` `((``1` `+ K * (sum)) == N) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `N1 = ``1570153``, K1 = ``12``; ` `    ``int` `N2 = ``321``, K2 = ``3``; ` `     `  `    ``// First two statements test  ` `    ``// against the condition  ` `    ``// N = 1 + K*(sum(proper divisors)) ` `    ``if` `(isHyperPerfect(N1, K1)) ` `        ``System.out.println (N1 + ``" is "` `+ K1 +  ` `                            ``"-HyperPerfect"` `); ` `    ``else` `        ``System.out.println(N1 + ``" is not "` `+ K1 +  ` `                               ``"-HyperPerfect"` `); ` `     `  `    ``if` `(isHyperPerfect(N2, K2)) ` `        ``System.out.println( N2 + ``" is "` `+ K2 +  ` `                            ``"-HyperPerfect"` `); ` `    ``else` `        ``System.out.println(N2 + ``" is not "` `+ K2 +  ` `                                ``"-HyperPerfect"``); ` `} ` `} ` ` `  `// This code is contributed by ajit `

## C#

 `// C# program to check  ` `// whether a given number ` `// is k-hyperperfect ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// function to find the  ` `// sum of all proper  ` `// divisors (excluding  ` `// 1 and N) ` `static` `int` `divisorSum(``int` `N,  ` `                      ``int` `K) ` `{ ` `    ``int` `sum = 0; ` ` `  `    ``// Iterate only until  ` `    ``// sqrt N as we are  ` `    ``// going to generate  ` `    ``// pairs to produce ` `    ``// divisors ` `    ``for` `(``int` `i = 2 ; ` `             ``i <= Math.Ceiling(Math.Sqrt(N));  ` `             ``i++) ` ` `  `        ``// As divisors occur in  ` `        ``// pairs, we can take  ` `        ``// the values i and N/i  ` `        ``// as long as i divides N ` `        ``if` `(N % i == 0) ` `            ``sum += (i + N / i); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Function to check  ` `// whether the given ` `// number is prime ` `static` `bool` `isPrime(``int` `n) ` `{ ` `    ``// base and corner cases ` `    ``if` `(n == 1 || n == 0) ` `        ``return` `false``; ` ` `  `    ``if` `(n <= 3) ` `        ``return` `true``; ` ` `  `    ``// Since integers can be  ` `    ``// represented as some  ` `    ``// 6*k + y where y >= 0,  ` `    ``// we can eliminate all  ` `    ``// integers that can be  ` `    ``// expressed in this form ` `    ``if` `(n % 2 == 0 ||  ` `        ``n % 3 == 0) ` `        ``return` `false``; ` ` `  `    ``// start from 5 as this  ` `    ``// is the next prime number ` `    ``for` `(``int` `i = 5;  ` `            ``i * i <= n; i = i + 6) ` `        ``if` `(n % i == 0 ||  ` `            ``n % ( i + 2 ) == 0) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Returns true if N is  ` `// a K-Hyperperfect number ` `// Else returns false. ` `static` `bool` `isHyperPerfect(``int` `N, ` `                           ``int` `K) ` `{ ` `    ``int` `sum = divisorSum(N, K); ` ` `  `    ``// Condition from the  ` `    ``// definition of hyperperfect ` `    ``if` `((1 + K * (sum)) == N) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver Code ` `static` `public` `void` `Main () ` `{ ` ` `  `int` `N1 = 1570153, K1 = 12; ` `int` `N2 = 321, K2 = 3; ` ` `  `// First two statements  ` `// test against the   ` `// condition N = 1 + K* ` `// (sum(proper divisors)) ` `if` `(isHyperPerfect(N1, K1)) ` `    ``Console.WriteLine(N1 + ``" is "` `+ K1 +  ` `                      ``"-HyperPerfect"` `); ` `else` `    ``Console.WriteLine(N1 + ``" is not "` `+ K1 +  ` `                          ``"-HyperPerfect"` `); ` ` `  `if` `(isHyperPerfect(N2, K2)) ` `    ``Console.WriteLine( N2 + ``" is "` `+ K2 +  ` `                       ``"-HyperPerfect"` `); ` `else` `    ``Console.WriteLine(N2 + ``" is not "` `+ K2 +  ` `                           ``"-HyperPerfect"``); ` `} ` `} ` ` `  `// This code is contributed ` `// by akt_mit `

## PHP

 `= 0, we can  ` `    ``// eliminate all integers that  ` `    ``// can be expressed in this form ` `    ``if` `(``\$n` `% 2 == 0 || ``\$n` `% 3 == 0) ` `        ``return` `false; ` ` `  `    ``// start from 5 as this ` `    ``// is the next prime number ` `    ``for` `(``\$i` `= 5;  ` `         ``\$i` `* ``\$i` `<= ``\$n``; ``\$i` `= ``\$i` `+ 6) ` `        ``if` `(``\$n` `% ``\$i` `== 0 ||  ` `            ``\$n` `% (``\$i` `+ 2) == 0) ` `            ``return` `false; ` ` `  `    ``return` `true; ` `} ` ` `  `// Returns true if N is a  ` `// K-Hyperperfect number ` `// Else returns false. ` `function` `isHyperPerfect(``\$N``, ``\$K``) ` `{ ` `    ``\$sum` `= divisorSum(``\$N``, ``\$K``); ` ` `  `    ``// Condition from the  ` `    ``// definition of hyperperfect ` `    ``if` `((1 + ``\$K` `* (``\$sum``)) == ``\$N``) ` `        ``return` `true; ` `    ``else` `        ``return` `false; ` `} ` ` `  `// Driver Code ` `\$N1` `= 1570153; ` `\$K1` `= 12; ` `\$N2` `= 321; ` `\$K2` `= 3; ` ` `  `// First two statements test  ` `// against the condition  ` `// N = 1 + K*(sum(proper divisors)) ` `if` `(isHyperPerfect(``\$N1``, ``\$K1``)) ` `    ``echo` `\$N1` `, ``" is "` `, ``\$K1``,  ` `           ``"-HyperPerfect"` `, ``"\n"``; ` `else` `    ``echo` `\$N1` `, ``" is not "` `, ``\$K1``,  ` `         ``"-HyperPerfect"` `, ``"\n"``; ` ` `  `if` `(isHyperPerfect(``\$N2``, ``\$K2``)) ` `    ``echo` `\$N2` `, ``" is "` `, K2,  ` `          ``"-HyperPerfect"` `, ``"\n"``; ` `else` `    ``echo` `\$N2` `, ``" is not "` `, ``\$K2` `, ` `          ``"-HyperPerfect"` `, ``"\n"``; ` ` `  `// This code is contributed ` `// by akt_mit ` `?> `

Output:

```1570153 is 12-HyperPerfect
321 is not 3-HyperPerfect
```

Given k, we can perform a few checks in special cases to determine whether the number is hyperperfect:

1. If K > 1 and K is odd , then let p = (3*k+1)/2 and q = 3*k+4 . If p and q are prime, then p2q is k-hyperperfect
2. If p and q are distinct odd primes such that K(p + q ) = pq – 1 for some positive integral value of K, then pq is k-hyperperfect

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Improved By : jit_t