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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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

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PHP

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<?php
// PHP implementation of the approach
  
// Function that returns true if n is prime
function 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 || $n % 3 == 0)
        return false;
  
    for ($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
function isDivisible($n)
{
    if (isPrime($n + 1))
        return false;
    return true;
}
  
// Driver code
$n = 6;
if (isDivisible($n))
    echo "Yes";
else
    echo "No";
  
// This code is contributed by Akanksha Rai
?>

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Output:

No


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