Competition · AMC preparation · step 4 of 4
AMC 8 · 2009 · #17
Grade 8 number-theoryPick an answer.
AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Factor 360 into primes
Factor 360 into primes to get 2³ · 3² · 5¹ — the 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.
6.NS.B.4Identify SubproblemsFind the smallest x
For a perfect square, raise each exponent to the next even number, so the missing factor is x = 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.
The smallest positive integer x that makes 360x a perfect square is the one that raises each prime exponent of 360 = 2³ · 3² · 5¹ up to the next even number, so x = 2¹ · 5¹.
▸ Why?
A whole number is a perfect square exactly when every prime in its factorization appears an even number of times, so the job is to turn the exponents 3, 2, 1 into even numbers.
▸ Why?
A perfect square is some whole number multiplied by itself, and writing that number's prime factors down a second time doubles how many times each prime appears, so a square always has even exponents; running this backward, even exponents split in half to rebuild the root.
▸ Why?
An exponent only records how many times a prime is used as a factor, so laying that same prime down again adds the identical count a second time and lands on twice the original, which is even.
▸ Why?
Multiplying 360 by x sets x's prime factors alongside 360's, so each prime's exponent in 360x is 360's exponent plus whatever exponent x supplies, and the smallest x adds only enough on each prime to reach the next even number.
▸ Why?
Since an exponent counts how many times a prime is written as a factor, gathering 360's copies of a prime together with x's copies makes the new count the sum of the two.
▸ Why?
The exponent on the prime 3 is already even, and adding no copies of it leaves that exponent unchanged, so x can leave that prime out entirely and stay as small as possible.
Check x equals 10
Check: 360 · 10 = 3600 = 60², a perfect square.
Confirming the square root explicitly is the quick sanity check that the exponent bookkeeping was right.
8.EE.A.2Identify SubproblemsFind the smallest y
For a perfect cube, raise each exponent to the next multiple of 3, so the missing factor is y = 75.
Same pattern as before, swapping "next even" for "next multiple of 3" — the rule is one line longer than the work.
8.EE.A.2Draw A Venn DiagramCheck y equals 75
Check: 360 · 75 = 2³ · 3³ · 5³ = 30³ = 27000, a perfect cube.
Seeing the matching exponents collapse into (2 · 3 · 5)³ is the satisfying visual that the cube condition is met.
8.EE.A.2Identify SubproblemsAdd x and y
Add the two subproblem answers: x + y = 10 + 75 = 85 → (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.
4.NBT.B.4Identify SubproblemsBreak 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.
- Factor 360 into primes
- Find the smallest x
- Check x equals 10
- Find the smallest y
- Check y equals 75
- Add x and y
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