Prove: for any natural number n, either n is prime, n is a perfect square, or n|(n-1)!

Suppose that a natural number n is neither a prime nor a perfect square. Then, there exist natural numbers p and q such that p < n, q < n , p ≠ q and pq = n.

Now, note that (n - 1)! = (n - 1)(n - 2)(n - 3)...(3)(2)(1).

Since p < n and q < n, p = (n - k) and q = (n - l) where k ≠ l and k, l ∈ {∈1, 2, 3, ... , n - 1}. But we know that (n - k)(n - l) | (n - 1)!

Hence, pq | (n - 1)! and therefore n | (n - 1)!

Hence, any natural number n is either prime, a complete square or divides (n - 1)!, as required.

a. the are of the rectangle is 18.25

b. 18.25 times two is 36.50

rae me : d

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.