AMC 10 · 2012 · #24
Grade 8 arithmeticPick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Each equation alone is a tangle of squares and cross-terms. Tool #15 (Organize Information in More Ways) is the key move: add the two equations. The nasty cross-terms line up so the sum collapses into the symmetric shape a² + b² + c² - ab - bc - ca, which is exactly half of (a-b)² + (b-c)² + (c-a)². That turns the problem into a statement about the gaps between the three numbers. Tool #4 (Introduce a Variable) names those gaps x = a-b and y = b-c; Tool #3 (Eliminate Possibilities) pins the gaps down because a small sum of squares has only a few whole-number solutions; Tool #7 (Identify Subproblems) then splits the finish into two cases and checks each against equation (1).
Add the two equations
Add the two equations term by term: the cross-terms partly cancel and the right side becomes 14. Halve both sides.
Two equations about the same unknowns can be added; here the ugly cross-terms partly cancel and leave a clean symmetric expression.
8.EE.C.8Organize Information In More WaysRewrite as a sum of squares
That symmetric form is exactly half of (a-b)² + (b-c)² + (c-a)², so multiply both sides by 2.
A symmetric quadratic in three variables repackages into squared differences, which measure how far apart the numbers are.
A symmetric quadratic in several unknowns repackages into squared differences.
▸ Why?
Opening each squared difference sends every piece against every piece, producing exactly those terms.
▸ Why?
Those squared gaps measure how far apart the numbers are, which is what the sum really encodes.
Name the gaps
Let x = a - b and y = b - c, non-negative integers with a - c = x + y; substituting and halving leaves x² + xy + y² = 7.
Working with the gaps instead of the numbers themselves shrinks three unknowns down to two small ones.
6.EE.B.6Introduce A VariablePin down the gaps
Each gap is at most 2, and the only non-negative pair with x² + xy + y² = 7 is 1 and 2 in some order, so a - c = 3.
A sum of squares equal to a small number has only a handful of whole-number solutions, so you can just list them.
6.EE.B.5Eliminate PossibilitiesTurn each case into an equation in a
Feed each case into equation (1): (a-1, a-3) collapses to 7a - 10 = 2011, and (a-2, a-3) to 8a - 13 = 2011.
Substituting the known gaps collapses the quadratic in (1) into a single linear equation in a.
6.EE.A.1Identify SubproblemsReject Case A
Case A needs 7a = 2021, but 2021 leaves remainder 5 on division by 7, so a is not an integer — drop it.
An integer unknown can only survive if the arithmetic divides evenly; a leftover remainder kills the case.
7.NS.A.2Eliminate PossibilitiesSolve Case B
Case B gives 8a = 2024, so a = 253 with b = 251, c = 250 — both original equations check out.
The one surviving case gives a clean linear equation whose whole-number solution is the answer.
8.EE.C.7Identify SubproblemsWhen two equations both look ugly, add them first — here the mess collapses into a sum of squared gaps, and once you know the numbers are only 1, 2, and 3 apart, one quick divisibility check leaves a = 253.
- Add the two equations
- Rewrite as a sum of squares
- Name the gaps
- Pin down the gaps
- Turn each case into an equation in a
- Reject Case A
- Solve Case B