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.
Input: N = 3
Product = 1 * 2 * 3 = 6
Sum = 1 + 2 + 3 = 6
Input: N = 6
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:
- Check whether factorial of N is divisible by sum of first N natural numbers
- Check if factorial of N is divisible by the sum of squares of first N natural numbers
- Sum of first N natural numbers which are divisible by X or Y
- Sum of first N natural numbers which are divisible by 2 and 7
- Number of pairs from the first N natural numbers whose sum is divisible by K
- Count natural numbers whose factorials are divisible by x but not y
- Count pairs of numbers from 1 to N with Product divisible by their Sum
- Sum and Product of all Composite numbers which are divisible by k in an array
- Check whether product of digits at even places of a number is divisible by K
- Check whether product of 'n' numbers is even or odd
- Check if all sub-numbers have distinct Digit product
- Check if the number formed by the last digits of N numbers is divisible by 10 or not
- Check whether product of digits at even places is divisible by sum of digits at odd place of a number
- Check if each element of the given array is the product of exactly K prime numbers
- Check if product of array containing prime numbers is a perfect square
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