# Multiply perfect number

A number N is said to be Multiply-perfect numbers if N divides sigma(N), where sigma(N) = sum of all divisors of N.

The first few Multiply-perfect numbers are:

1, 6, 28, 120, 496, 672, ……..

### Check if N is a Multiply-perfect number

Given a number N, the task is to find if this number is Multiply-perfect number or not.

Examples:

Input: N = 120
Output: YES
Explanation:
Sum of 120’s divisors is 1 + 2 + 3 + 4 + 5 + 6 + 8 + 10 + 12 + 15 + 20 + 24 + 30 + 40 + 60 + 120 = 360 and 120 divides 360.
Therefore, 120 is a Multiply-perfect number.

Input: N = 32
Output: No

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

Approach: For a number N to be Multiply-perfect number, the following condition should hold true: sigma(N) % N = 0, where sigma(N) = sum of all divisors of N. Therefore, we will find sum of all divisors of N and check if it is divisible by N or not. If divisible 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 find the ` `// 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; ` ` `  `            ``// Otherwise take both ` `            ``else` `{ ` `                ``sum = sum + i; ` `                ``sum = sum + (n / i); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Function to check Multiply-perfect number ` `bool` `MultiplyPerfectNumber(``int` `n) ` `{ ` `    ``if` `(getSum(n) % n == 0) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `n = 28; ` `    ``if` `(MultiplyPerfectNumber(n)) { ` `        ``cout << ``"Yes"``; ` `    ``} ` `    ``else` `{ ` `        ``cout << ``"No"``; ` `    ``} ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `class` `GFG{ ` `     `  `// Function to find the ` `// 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 Multiply-perfect number ` `static` `boolean` `MultiplyPerfectNumber(``int` `n) ` `{ ` `    ``if` `(getSum(n) % n == ``0``) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``28``; ` `     `  `    ``if` `(MultiplyPerfectNumber(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 find the ` `# sum of divisors ` `def` `getSum(n): ` ` `  `    ``sum1 ``=` `0``; ` ` `  `    ``# Note that this loop ` `    ``# runs till square root of N ` `    ``for` `i ``in` `range``(``1``, ``int``(math.sqrt(n))): ` `        ``if` `(n ``%` `i ``=``=` `0``): ` ` `  `            ``# If divisors are equal, ` `            ``# take only one of them ` `            ``if` `(n ``/``/` `i ``=``=` `i): ` `                ``sum1 ``=` `sum1 ``+` `i; ` ` `  `            ``# Otherwise take both ` `            ``else``: ` `                ``sum1 ``=` `sum1 ``+` `i; ` `                ``sum1 ``=` `sum1 ``+` `(n ``/``/` `i); ` `             `  `    ``return` `sum1; ` ` `  `# Function to check Multiply-perfect number ` `def` `MultiplyPerfectNumber(n): ` ` `  `    ``if` `(getSum(n) ``%` `n ``=``=` `0``): ` `        ``return` `True``; ` `    ``else``: ` `        ``return` `False``; ` ` `  `# Driver code ` `n ``=` `28``; ` `if` `(MultiplyPerfectNumber(n)): ` `    ``print``(``"Yes"``); ` `else``: ` `    ``print``(``"No"``); ` ` `  `# This code is contributed by Code_Mech `

## C#

 `// C# implementation of the above approach ` `using` `System; ` `class` `GFG{ ` `     `  `// Function to find the ` `// 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 Multiply-perfect number ` `static` `bool` `MultiplyPerfectNumber(``int` `n) ` `{ ` `    ``if` `(getSum(n) % n == 0) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 28; ` `     `  `    ``if` `(MultiplyPerfectNumber(n)) ` `    ``{ ` `        ``Console.Write(``"Yes"``); ` `    ``} ` `    ``else` `    ``{ ` `        ``Console.Write(``"No"``); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```Yes
```

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Improved By : btc_148, Code_Mech