AMC 10 · 2025 · #10
Grade 8 algebraPick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
A ratio of two ugly cubics looks hopeless head-on, so Tool #7 (Identify Subproblems) splits it into three clean jobs: factor f, factor g, then compare. Tool #6 (Guess and Check) cracks each cubic by testing small integers to find a first root. Once both are factored, Tool #15 (Organize Information in More Ways) rewrites the reduced fraction (n-2)/(n-3) as 1+1/(n-3), which turns "when is this an integer?" into a simple divisibility question. Tool #3 (Eliminate Possibilities) then removes the value that secretly breaks the original fraction.
Factor the top cubic
Testing n=-1 in f gives 0, so n+1 is a factor; dividing it out leaves n²-6n+8, hence f(n)=(n+1)(n-2)(n-4).
A polynomial that equals zero at some number must have that number's factor hiding inside it.
A polynomial that equals zero at some number must have that number's factor hiding inside it.
▸ Why?
A product is zero only where one of its factors is zero, so the root marks a factor.
▸ Why?
Dividing by that factor leaves no remainder, which is what makes the factoring exact.
Factor the bottom cubic
Same trick on g: n=-1 works again, and the leftover quadratic is n²-7n+12, so g(n)=(n+1)(n-3)(n-4).
Breaking each cubic into three simple factors makes the two expressions easy to line up side by side.
7.EE.A.1Identify SubproblemsCancel the shared factors
Both share n+1 and n-4, so cancelling them — legal only where they are nonzero — leaves just (n-2)/(n-3).
Matching factors on top and bottom divide to 1, so the scary cubics collapse into one tidy ratio.
7.NS.A.2Identify SubproblemsSplit off the whole part
Write n-2 as (n-3)+1, and the ratio splits into 1+1/(n-3), so only the leftover piece decides the outcome.
Peeling off the built-in 1 leaves just one tiny fraction to worry about.
6.EE.A.3Organize Information In More WaysWhen is the leftover a whole number?
1/(n-3) is whole only if n-3 divides 1, and the only integer divisors of 1 are 1 and -1.
A unit fraction 1/k is only whole when its denominator is as small as it can be: ±1.
4.OA.B.4Organize Information In More WaysSolve for the candidate values
Solve the two tiny equations: n-3=1 gives n=4 and n-3=-1 gives n=2, so the candidates are n=4 and n=2.
Each allowed value of n-3 points straight to one value of n.
8.EE.C.7Guess And CheckThrow out the fake solution and add up
At n=4 both cubics vanish, so 0/0 is undefined and n=4 is out; only n=2 survives, making the sum 2, choice (A).
Cancelling can hide a spot where the real fraction blows up, so always test the survivors in the original.
6.EE.A.2Eliminate PossibilitiesFactor both cubics, cancel what they share to get 1+1/(n-3), and the only whole-number results come from n-3=±1 — but toss out n=4 because it makes the real fraction 0/0.
- Factor the top cubic
- Factor the bottom cubic
- Cancel the shared factors
- Split off the whole part
- When is the leftover a whole number?
- Solve for the candidate values
- Throw out the fake solution and add up