AMC 8 · 2009 · #17
Grade 8 number-theoryProblem
The positive integers x and y are the two smallest positive integers for which the product of 360 and x is a square and the product of 360 and y is a cube. What is the sum of x and y?
Pick an answer.
AMC 8 2009 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: Let $x$ be the smallest positive integer that makes $360x$ a perfect square, and let $y$ be the smallest positive integer that makes $360y$ a perfect cube. Find $x + y$.
Givens: Base number is $360$; $360 \cdot x$ must be a perfect square; $360 \cdot y$ must be a perfect cube; $x$ and $y$ are the smallest positive integers with these properties; Answer choices: (A) $80$, (B) $85$, (C) $115$, (D) $165$, (E) $610$
Unknowns: The sum $x + y$
Understand
Restated: Let $x$ be the smallest positive integer that makes $360x$ a perfect square, and let $y$ be the smallest positive integer that makes $360y$ a perfect cube. Find $x + y$.
Givens: Base number is $360$; $360 \cdot x$ must be a perfect square; $360 \cdot y$ must be a perfect cube; $x$ and $y$ are the smallest positive integers with these properties; Answer choices: (A) $80$, (B) $85$, (C) $115$, (D) $165$, (E) $610$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #12 Find a Pattern
The problem looks like one question but is really two independent subproblems glued together by a final sum, so Tool #7 (Identify Subproblems) is the natural opener: find $x$, find $y$, then add. Both subproblems share the same setup: factor $360$ into primes and look at the exponents. Tool #12 (Find a Pattern) names the pattern that runs through both — to make a number a perfect square (or cube), each prime exponent has to be bumped up to the next even number (or next multiple of $3$). That single rule mechanically gives both $x$ and $y$ from the factorization of $360$.
Execute — Answer: B
6.NS.B.4 Step 1 - Factor $360$ into primes.
- Pull out $10 = 2 \cdot 5$ and $36 = 2^2 \cdot 3^2$ from $360 = 36 \cdot 10$, then combine: $360 = 2^3 \cdot 3^2 \cdot 5^1$.
- The three prime exponents are $3, 2, 1$.
💡 Reading off prime exponents is the Grade 6 number-theory move that turns a square/cube question into a question about each exponent separately.
8.EE.A.2 Step 2 - Find the smallest $x$ so that $360x$ is a perfect square.
- A perfect square has every prime exponent even.
- Starting from exponents $(3, 2, 1)$ for primes $(2, 3, 5)$, bump each up to the next even number: $3 \to 4$, $2 \to 2$, $1 \to 2$.
- So $x$ must supply $2^{4-3} \cdot 3^{2-2} \cdot 5^{2-1} = 2^1 \cdot 3^0 \cdot 5^1$.
💡 "Push every exponent up to the next even number" is the pattern (Tool #12) that defines a perfect square via prime factorization — exactly the Grade 8 reasoning that connects exponents to squares.
8.EE.A.2 Step 3 - Check $x = 10$.
- Multiplying $360 \cdot 10 = 3600$, and $60^2 = 3600$, so $360 \cdot 10$ really is a perfect square.
💡 Confirming the square root explicitly is the quick sanity check that the exponent bookkeeping was right.
8.EE.A.2 Step 4 - Find the smallest $y$ so that $360y$ is a perfect cube.
- A perfect cube has every prime exponent divisible by $3$.
- Push exponents $(3, 2, 1)$ up to the next multiple of $3$: $3 \to 3$, $2 \to 3$, $1 \to 3$.
- So $y$ must supply $2^{3-3} \cdot 3^{3-2} \cdot 5^{3-1} = 2^0 \cdot 3^1 \cdot 5^2$.
💡 Same pattern as before, swapping "next even" for "next multiple of $3$" — the rule is one line longer than the work.
8.EE.A.2 Step 5 - Check $y = 75$.
- Then $360 \cdot 75 = 2^3 \cdot 3^2 \cdot 5 \cdot 3 \cdot 5^2 = 2^3 \cdot 3^3 \cdot 5^3 = (2 \cdot 3 \cdot 5)^3 = 30^3 = 27000$.
- So $360 \cdot 75$ is indeed a perfect cube.
💡 Seeing the matching exponents collapse into $(2 \cdot 3 \cdot 5)^3$ is the satisfying visual that the cube condition is met.
4.NBT.B.4 Step 6 Add the two answers from the subproblems to finish.
💡 The final "and now add" step is the close of the Tool #7 split — once both subproblems are solved, the rest is one-digit addition.
6.NS.B.4 Factor $360$ into primes. Pull out $10 = 2 \cdot 5$ and $36 = 2^2 \cdot 3^2$ fro 8.EE.A.2 Find the smallest $x$ so that $360x$ is a perfect square. A perfect square has e 8.EE.A.2 Check $x = 10$. Multiplying $360 \cdot 10 = 3600$, and $60^2 = 3600$, so $360 \c 8.EE.A.2 Find the smallest $y$ so that $360y$ is a perfect cube. A perfect cube has every 8.EE.A.2 Check $y = 75$. Then $360 \cdot 75 = 2^3 \cdot 3^2 \cdot 5 \cdot 3 \cdot 5^2 = 2 4.NBT.B.4 Add the two answers from the subproblems to finish. Review
Reasonableness: Sanity-check the sizes. The answer choices range from $80$ to $610$, and most of them ($85$, $115$, $165$) are in the same ballpark as $10 + 75$. The two extreme choices $80$ and $610$ would force $y$ to be near $70$ or near $600$ — but $360 \cdot 600 = 216000$ is exactly $60^3$, not the smallest cube multiplier, and $360 \cdot 70$ is $25200$, not a cube. The compact value $85$ matches the prime-factorization argument, so (B) is consistent with both the structure and the size of the choices.
Alternative: Tool #6 (Guess and Check), working through the choices. The pairs $(x, y)$ must satisfy $x + y \in \{80, 85, 115, 165, 610\}$ with $x = 10$ forced (the only smaller candidate $x = 1$ gives $360$, not a square; $x = 2, 5$ also fail). With $x = 10$ pinned, $y = 85 - 10 = 75$ is the only choice that gives $360y = 27000 = 30^3$; the other differences $70, 105, 155, 600$ all fail the cube test. So (B) is the unique survivor.
CCSS standards used (min grade 8)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing the final sum $x + y = 10 + 75 = 85$ once both subproblems are solved.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Carrying out the prime factorization $360 = 2^3 \cdot 3^2 \cdot 5^1$ that the whole solution rests on.)8.EE.A.2Use square root and cube root symbols and represent solutions; know perfect squares and perfect cubes (Translating "$360x$ is a perfect square" and "$360y$ is a perfect cube" into conditions on prime exponents (all even, or all multiples of $3$) and verifying $3600 = 60^2$ and $27000 = 30^3$.)
⭐ Break $360$ into primes, then bump each exponent up to the next even number for squares and the next multiple of $3$ for cubes — that single pattern hands you both $x$ and $y$.
⭐ Break $360$ into primes, then bump each exponent up to the next even number for squares and the next multiple of $3$ for cubes — that single pattern hands you both $x$ and $y$.
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