Consider the set S of all four-digit natural numbers greater than 9000, each having exactly 16 factors. If the prime factorization of each element in S is the product of powers of only two distinct prime numbers, the sum of all elements in S is________

To solve this problem, we need to understand the concept of prime factorization and the number of factors.

1. A four-digit number greater than 9000 having exactly 16 factors means that the number is either in the form of p^15 (where p is a prime number) or in the form of p^3*q^3 (where p and q are distinct prime numbers).

2. The first case (p^15) is not possible because the smallest prime number is 2 and 2^15 is already a five-digit number.

3. So, we consider the second case (p^3*q^3). The smallest primes are 2 and 3, but 2^3 * 3^3 is less than 9000. The next prime is 5, but 2^3 * 5^3 is still less than 9000. The next prime is 7, and 2^3 * 7^3 is greater than 9000. So, the smallest possible values for p and q are 2 and 7.

4. We also need to consider the case where p and q are swapped (i.e., p=7 and q=2), but this gives us the same number, so we don't count it twice.

5. Therefore, the only four-digit number greater than 9000 that is the product of powers of two distinct prime numbers and has exactly 16 factors is 2^3 * 7^3 = 6860.

6. The sum of all elements in S is therefore 6860.