AMC 8 · 2009 · #17

Grade 8 number-theory

Archetype Arithmetic Expression Decomposition

prime-factorizationperfect-squaresexponents identify-subproblemspattern-recognition ↑ Prerequisites: prime-factorizationexponents

📏 Medium solution 💡 3 insights

Problem

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.

(A)

(B)

(C)

115

(D)

165

(E)

610

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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 $x + y$.

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

#7 Identify Subproblems 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 $3, 2, 1$.

$$360 = 36 \cdot 10 = (2^2 \cdot 3^2)(2 \cdot 5) = 2^3 \cdot 3^2 \cdot 5^1$$

💡 Reading off prime exponents is the Grade 6 number-theory move that turns a square/cube question into a question about each exponent separately.

#12 Find a Pattern 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 $2^{4-3} \cdot 3^{2-2} \cdot 5^{2-1} = 2^1 \cdot 3^0 \cdot 5^1$.

$$x = 2^1 \cdot 3^0 \cdot 5^1 = 2 \cdot 5 = 10$$

💡 "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.

#7 Identify Subproblems 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.

$$360 \cdot 10 = 3600 = 60^2 \;\checkmark$$

💡 Confirming the square root explicitly is the quick sanity check that the exponent bookkeeping was right.

#12 Find a Pattern 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 $2^{3-3} \cdot 3^{3-2} \cdot 5^{3-1} = 2^0 \cdot 3^1 \cdot 5^2$.

$$y = 2^0 \cdot 3^1 \cdot 5^2 = 3 \cdot 25 = 75$$

💡 Same pattern as before, swapping "next even" for "next multiple of $3$" — the rule is one line longer than the work.

#7 Identify Subproblems 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.

$$360 \cdot 75 = 2^3 \cdot 3^3 \cdot 5^3 = 30^3 = 27000 \;\checkmark$$

💡 Seeing the matching exponents collapse into $(2 \cdot 3 \cdot 5)^3$ is the satisfying visual that the cube condition is met.

#7 Identify Subproblems 4.NBT.B.4 Step 6

Add the two answers from the subproblems to finish.

$$x + y = 10 + 75 = 85 \;\Rightarrow\; \textbf{(B)}$$

💡 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.

[1] #7 6.NS.B.4 Factor $360$ into primes. Pull out $10 = 2 \cdot 5$ and $36 = 2^2 \cdot 3^2$ fro

[2] #12 8.EE.A.2 Find the smallest $x$ so that $360x$ is a perfect square. A perfect square has e

[3] #7 8.EE.A.2 Check $x = 10$. Multiplying $360 \cdot 10 = 3600$, and $60^2 = 3600$, so $360 \c

[4] #12 8.EE.A.2 Find the smallest $y$ so that $360y$ is a perfect cube. A perfect cube has every

[5] #7 8.EE.A.2 Check $y = 75$. Then $360 \cdot 75 = 2^3 \cdot 3^2 \cdot 5 \cdot 3 \cdot 5^2 = 2

[6] #7 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.4 Fluently add and subtract multi-digit whole numbers (Computing the final sum $x + y = 10 + 75 = 85$ once both subproblems are solved.)
6.NS.B.4 Find 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.2 Use 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 $y$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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