Question

Three positive integers are each greater than $1,$ have a product of $27000 ,$ and are pairwise relatively prime. What is their sum?

$100$

$137$

$156$

$160$

$165$

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 .