The best starting point for this question is to do some trial and error.
3 is not a factor of 2!
4 is not a factor of 3!
5 is not a factor of 4!
6 is a factor of 5!
7 is not a factor of 6!
8 is a factor of 7!
The first thing we see is that n + 1 cannot be prime. If (n + 1) were prime, it cannot be a factor of n!.
So, we can eliminate all primes.
Now, let us think of all numbers where (n + 1) is not prime. In this instance, we should be able to write (n + 1) as a * b where a,b are not 1 and (n + 1). So, (a, b) will lie in the set {1, 2, 3........n} or, a * b will be a factor of (n + 1)!
So, for any composite (n + 1), (n + 1) will always be a factor of n! {Is there any exception?}
For any prime number (n + 1), (n + 1) will never be a factor of n!
The above rule works well even for all the examples we have seen, except when (n + 1) = 4. 4 = 2 * 2; So, 4 is not a factor of 3!. But this is the only exception.
Counting on from here, we can see that n can take values 5, 7, 8, 9, 11, 13, 14, 15, 17, 19, 20, 21, 23, 24, 25, 26, 27, 29. Essentially, all numbers where N + 1 is greater than 4 and is not prime will feature in this list. If N + 1 is not prime, we should be able to write it as a product of 2 numbers less than N + 1. This will feature in n!. This question is just a different way of asking one to count primes (and then account for the exception of 4)
n can take 18 different values.
Correct Answer: 18 different values
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