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 = 6Input: 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:

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