AMC 10 · 2013 · #9

Grade 6 number-theory
prime-factorizationgcdexponents identify-subproblemscasework ↑ Prerequisites: prime-factorization
📏 Medium solution 💡 2 insights
Problem
Three whole numbers are each bigger than 1, multiply to 27000, and share no common factor in any pair. What is the sum of the three numbers?

Pick an answer.

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

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Tool #7 (Identify Subproblems) splits the task into three clean jobs: break 27000 into prime powers, decide how those primes get grouped into three numbers, then add. The middle job is where Tool #3 (Eliminate Possibilities) does the heavy lifting: the pairwise-coprime rule forbids two numbers from sharing any prime, so a prime's full power cannot be split across two factors. That single rule eliminates every grouping except one, so no guessing is needed.

1STEP 1

Prime factorize 27000

Since 27000 = 27 × 1000, the prime factorization is 2³ × 3³ × 5³ — only the primes 2, 3, and 5 appear.

27000 = 27 × 1000 = 3³ × (2 × 5)³ = 2³ × 3³ × 5³
2STEP 2

Coprime rule forces the grouping

Splitting 2³ would leave two even numbers sharing the factor 2, so each prime block stays whole: one number each.

gcd(a,b)=gcd(b,c)=gcd(a,c)=1 → {2³, 3³, 5³}
3STEP 3

Compute the three numbers

Evaluating the blocks gives 8, 27, and 125, all above 1, and 8 × 27 × 125 = 27000 confirms the product.

2³ = 8, 3³ = 27, 5³ = 125
4STEP 4

Add the three numbers

Adding them gives 8 + 27 + 125 = 160, which only choice (D) matches, so the rest are eliminated.

8 + 27 + 125 = 160 → (D)
Answer
160
The three numbers 8, 27, 125 are each above 1, multiply to 8 × 27 × 125 = 27000, and are pairwise coprime because they are powers of different primes. Their sum 160 lands squarely among the choices as (D). The near-miss (E) 165 would need a slightly larger set, but no other coprime split of 27000 into three factors above 1 exists, so 160 is the only possible sum.
💡Key takeaway

When numbers must be pairwise relatively prime, each prime's full power has to stay inside one number — so factor into primes, hand out one prime block per number, then add.

  • Prime factorize 27000
  • Coprime rule forces the grouping
  • Compute the three numbers
  • Add the three numbers