# Check if product of first N natural numbers is divisible by their sum

Given an integer N, the task is to check whether the product of first N natural numbers is divisible by the sum of first N natural numbers.

Examples:

Input: N = 3
Output: Yes
Product = 1 * 2 * 3 = 6
Sum = 1 + 2 + 3 = 6

Input: N = 6
Output: No

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

Naive Approach: Find the sum and product of first N natural numbers and check whether the product is divisible by the sum.

Efficient Approach: We know that the sum and product of first N naturals are sum = (N * (N + 1)) / 2 and product = N! respectively. Now to check whether the product is divisible by the sum, we need to check if the remainder of the following equation is 0 or not.

N! / (N *(N + 1) / 2)
2 * (N – 1)! / N + 1
i.e. every factor of (N + 1) should be in (2 * (N – 1)!). So, if (N + 1) is a prime then we are sure that the product is not divisible by the sum.
So ultimately just check if (N + 1) is prime or not.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if n is prime ` `bool` `isPrime(``int` `n) ` `{ ` `    ``// Corner cases ` `    ``if` `(n <= 1) ` `        ``return` `false``; ` `    ``if` `(n <= 3) ` `        ``return` `true``; ` ` `  `    ``// This is checked so that we can skip ` `    ``// middle five numbers in below loop ` `    ``if` `(n % 2 == 0 || n % 3 == 0) ` `        ``return` `false``; ` ` `  `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6) ` `        ``if` `(n % i == 0 || n % (i + 2) == 0) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Function that return true if the product ` `// of the first n natural numbers is divisible ` `// by the sum of first n natural numbers ` `bool` `isDivisible(``int` `n) ` `{ ` `    ``if` `(isPrime(n + 1)) ` `        ``return` `false``; ` `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 6; ` `    ``if` `(isDivisible(n)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `// Function that returns true if n is prime ` `static` `boolean` `isPrime(``int` `n) ` `{ ` `    ``// Corner cases ` `    ``if` `(n <= ``1``) ` `        ``return` `false``; ` `    ``if` `(n <= ``3``) ` `        ``return` `true``; ` ` `  `    ``// This is checked so that we can skip ` `    ``// middle five numbers in below loop ` `    ``if` `(n % ``2` `== ``0` `|| n % ``3` `== ``0``) ` `        ``return` `false``; ` ` `  `    ``for` `(``int` `i = ``5``; i * i <= n; i = i + ``6``) ` `        ``if` `(n % i == ``0` `|| n % (i + ``2``) == ``0``) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Function that return true if the product ` `// of the first n natural numbers is divisible ` `// by the sum of first n natural numbers ` `static` `boolean` `isDivisible(``int` `n) ` `{ ` `    ``if` `(isPrime(n + ``1``)) ` `        ``return` `false``; ` `    ``return` `true``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``6``; ` `    ``if` `(isDivisible(n)) ` `        ``System.out.println(``"Yes"``); ` `    ``else` `        ``System.out.println(``"No"``); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech. `

## Python3

 `# Python 3 implementation of the approach ` `from` `math ``import` `sqrt ` ` `  `# Function that returns true if n is prime ` `def` `isPrime(n): ` `     `  `    ``# Corner cases ` `    ``if` `(n <``=` `1``): ` `        ``return` `False` `    ``if` `(n <``=` `3``): ` `        ``return` `True` ` `  `    ``# This is checked so that we can skip ` `    ``# middle five numbers in below loop ` `    ``if` `(n ``%` `2` `=``=` `0` `and` `n ``%` `3` `=``=` `0``): ` `        ``return` `False` ` `  `    ``for` `i ``in` `range``(``5``, ``int``(sqrt(n)) ``+` `1``, ``6``): ` `        ``if` `(n ``%` `i ``=``=` `0` `and` `n ``%` `(i ``+` `2``) ``=``=` `0``): ` `            ``return` `False` ` `  `    ``return` `True` ` `  `# Function that return true if the product ` `# of the first n natural numbers is divisible ` `# by the sum of first n natural numbers ` `def` `isDivisible(n): ` `    ``if` `(isPrime(n ``+` `1``)): ` `        ``return` `False` `    ``return` `True` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `6` `    ``if` `(isDivisible(n)): ` `        ``print``(``"Yes"``) ` `    ``else``: ` `        ``print``(``"No"``) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function that returns true if n is prime ` `static` `bool` `isPrime(``int` `n) ` `{ ` `    ``// Corner cases ` `    ``if` `(n <= 1) ` `        ``return` `false``; ` `    ``if` `(n <= 3) ` `        ``return` `true``; ` ` `  `    ``// This is checked so that we can skip ` `    ``// middle five numbers in below loop ` `    ``if` `(n % 2 == 0 || n % 3 == 0) ` `        ``return` `false``; ` ` `  `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6) ` `        ``if` `(n % i == 0 || n % (i + 2) == 0) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Function that return true if the product ` `// of the first n natural numbers is divisible ` `// by the sum of first n natural numbers ` `static` `bool` `isDivisible(``int` `n) ` `{ ` `    ``if` `(isPrime(n + 1)) ` `        ``return` `false``; ` `    ``return` `true``; ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `n = 6; ` `    ``if` `(isDivisible(n)) ` `        ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``); ` ` `  `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

 ` `

Output:

```No
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.