AMC 10 · 2013 · #9
Grade 6 number-theoryThree positive integers are each greater than 1, have a product of 27000, and are pairwise relatively prime. What is their sum?
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Find three whole numbers, each bigger than $1$, that multiply to $27000$ and share no common factor in any pair. Then add the three numbers together.
Givens: Three positive integers, each greater than $1$; Their product is $27000$; The numbers are pairwise relatively prime (any two of them share no factor except $1$); Answer choices: (A) $100$, (B) $137$, (C) $156$, (D) $160$, (E) $165$
Unknowns: The sum of the three numbers
Understand
Restated: Find three whole numbers, each bigger than $1$, that multiply to $27000$ and share no common factor in any pair. Then add the three numbers together.
Givens: Three positive integers, each greater than $1$; Their product is $27000$; The numbers are pairwise relatively prime (any two of them share no factor except $1$); Answer choices: (A) $100$, (B) $137$, (C) $156$, (D) $160$, (E) $165$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
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.
Execute — Answer: D
4.OA.B.4 Step 1 Prime factorize 27000
- Break the product into its prime building blocks.
- Since $27000 = 27 \times 1000$, and $27 = 3^3$ while $1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3$, the full factorization is $27000 = 2^3 \times 3^3 \times 5^3$.
- Only the primes $2$, $3$, and $5$ appear.
💡 Every whole number breaks into primes in exactly one way, so factoring reveals all the pieces you have to work with.
6.NS.B.4 Step 2 Coprime rule forces the grouping
- Two numbers are relatively prime only if they share no prime factor.
- So the whole block of $2$s must sit inside a single number — if $2^3$ were split, say $2$ in one number and $2^2$ in another, those two numbers would both be even and share the factor $2$, breaking the rule.
- The same is true for the $3$s and the $5$s.
- That gives exactly three prime-power blocks $2^3$, $3^3$, $5^3$ for the three numbers, one block each.
💡 If a prime were spread across two numbers, they would share that prime, so each prime must stay whole in one number.
6.EE.A.1 Step 3 Compute the three numbers
- Evaluate each prime power.
- $2^3 = 8$, $3^3 = 27$, and $5^3 = 125$.
- Each is greater than $1$, and as a check $8 \times 27 \times 125 = 27000$, matching the required product.
💡 A power like $5^3$ just means multiplying the base by itself three times.
4.NBT.B.4 Step 4 Add the three numbers
- Add the numbers to get the sum: $8 + 27 + 125 = 160$.
- Scanning the choices, only (D) equals $160$, so every other option is eliminated.
- The answer is (D).
💡 Once the three numbers are pinned down, the sum is a single addition.
4.OA.B.4 Break the product into its prime building blocks. Since $27000 = 27 \times 1000$ 6.NS.B.4 Two numbers are relatively prime only if they share no prime factor. So the whol 6.EE.A.1 Evaluate each prime power. $2^3 = 8$, $3^3 = 27$, and $5^3 = 125$. Each is great 4.NBT.B.4 Add the numbers to get the sum: $8 + 27 + 125 = 160$. Scanning the choices, only Review
Reasonableness: The three numbers $8$, $27$, $125$ are each above $1$, multiply to $8 \times 27 \times 125 = 27000$, 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.
Alternative: Try to build a different valid triple and see it is impossible. Any factor above $1$ must contain a whole prime block, and there are exactly three blocks ($2^3$, $3^3$, $5^3$) for three factors, so each factor gets exactly one block. Merging two blocks (like $8 \times 27 = 216$) would leave only two factors above $1$, not three. This confirms $\{8, 27, 125\}$ is forced and the sum must be $160$.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring $27000$ into the prime powers $2^3 \times 3^3 \times 5^3$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Using the pairwise-coprime (gcd $=1$) condition to keep each prime power inside a single number.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating $2^3 = 8$, $3^3 = 27$, and $5^3 = 125$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding $8 + 27 + 125 = 160$ for the final sum.)
⭐ 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.
⭐ 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.
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