AMC 10 · 2012 · #21
Grade 8 number-theoryalgebraPick an answer.
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
Adding the equations simplifies everything.
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
It rewrites as a sum of squared gaps.
A symmetric quadratic in three variables repackages into squared differences, which measure how far apart the numbers are.
A symmetric quadratic in three variables repackages into squared differences, which measure how far apart the numbers are.
▸ Why?
Expanding a squared difference spreads the multiplication out into two squares and a cross term.
▸ Why?
Those cross terms are exactly what the symmetric expression carries, so the two forms match term for term.
Name the gaps
Naming the gaps leaves one small equation.
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
Only two gap pairs are possible.
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
Each becomes a plain linear equation.
Substituting the known gaps collapses the quadratic in (1) into a single linear equation in a.
6.EE.A.1Identify SubproblemsReject Case A
One fails to give a whole number.
An integer unknown can only survive if the arithmetic divides evenly; a leftover remainder kills the case.
7.NS.A.2Eliminate PossibilitiesSolve Case B
The other gives 253, choice (E).
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