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2013 AMC 10B Problem 9

2013 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 2 of 5 · about 2 min · position P1-10

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

A)
$100$
B)
$137$
C)
$156$
D)
$160$
E)
$165$

Answer

Key insight

Pairwise relatively prime factors cannot share a prime, so each prime power of 27000 = 2^3 3^3 5^3 goes whole to one number: 8, 27, 125.

Solution

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 $\boxed{\textbf{(D)}\ 160}$ .

Why this works

"Pairwise relatively prime" means the prime factorizations are disjoint. So a coprime factorization of $n$ is a partition of the set of prime powers in $n$ ; with three primes and three factors greater than $1$ , the partition is forced. Always start such problems with the full prime factorization.

The trap

Splitting the factorization by size (e.g. 30, 30, 30) and ignoring the relatively-prime condition.

Common mistakes

Splitting the factorization by size (e.g. 30, 30, 30) and ignoring the relatively-prime condition.
Using $2, 3, 5$ or $2^3 \cdot 3, \ldots$ combinations that do not multiply to $27000$ ; check the product.

Techniques

Set up the equation/formula and compute; no special trick needed