2013 AMC 10B Problem 9

1. Break $27000$ into primes: $27000 = 27 \times 1000$, and $27 = 3 \times 3 \times 3$, $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$. So
$$27000 = 2^3 \times 3^3 \times 5^3.$$
2. "Pairwise relatively prime" means no two of the numbers share a common factor bigger than $1$. So if one number contains a $2$, no other number can contain a $2$. All three $2$s must go into the same number. The same holds for the $3$s and the $5$s.
3. That means each number is made of whole blocks from $2^3 = 8$, $3^3 = 27$, $5^3 = 125$.
4. There are three blocks and three numbers, and each number must be bigger than $1$. If one number took two blocks, another would get none and would equal $1$. So each number gets exactly one block: the numbers are $8$, $27$ and $125$.
5. Check: $8 \times 27 \times 125 = (8 \times 125) \times 27 = 1000 \times 27 = 27000$.
6. The sum is $8 + 27 + 125 = 160$.

The answer is $\boxed{\textbf{(D)}\ 160}$.

"Relatively prime" is a statement about prime factors, so start with the prime factorization. Once no prime may be shared, each prime's whole block has to sit in one number, and here the three blocks exactly fill the three numbers.

Factor: $27000 = 27 \cdot 1000 = 3^3 \cdot 2^3 \cdot 5^3$.

If two of the numbers shared a prime factor they would not be relatively prime, so every copy of a given prime must sit inside the same number. The three primes $2, 3, 5$ must therefore be distributed with all of $2^3$ in one number, all of $3^3$ in another, and all of $5^3$ in a third. Because each number exceeds $1$, no number can be left empty, so the numbers are exactly
$$2^3 = 8, \qquad 3^3 = 27, \qquad 5^3 = 125.$$

Their sum is $8 + 27 + 125 = 160$.

The answer is $\boxed{\textbf{(D)}\ 160}$.