Solution:

Only one of them is a multiple of $2,$ only one of them is a multiple of $3,$ and only one of them is a multiple of $5.$

Since each of the positive integers is greater than $1,$ each of them must be a multiple of one of the given primes.

Therefore, since $$27000=2^3\cdot 3^3\cdot 5^3,$$ the numbers must be $2^3,3^3,5^3,$ making their sum $160.$

Thus, the correct answer is D .