AMC 10 · 2020 · #3
Grade 7 algebraPick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #5 (Look for a Pattern): every numerator (a-3), (b-4), (c-5) is the negative of some denominator -(3-a), -(4-b), -(5-c). That single repeated structure is the whole problem. Tool #15 (Reorganize) — re-pair the fractions so each numerator sits over its own negative: · · . Tool #6 (Guess and Check) is an even faster sanity path: pick easy numbers (a=b=c=0) and just compute. Tool #3 (Eliminate) confirms — the expression must be a constant (no a,b,c left), so (C), (D), (E) all involve abc and are out.
Each numerator is the opposite of a denominator: (a-3) = -(3-a), same for the others.
Reversing the order in a subtraction flips its sign — Grade 6 understanding of opposite (negative) numbers.
6.NS.C.5Look For A PatternReorder the factors so each numerator sits over its own opposite denominator.
Commutative property of multiplication — change the pairing to expose the pattern.
3.OA.B.5Organize Information In More WaysEach re-paired fraction is = -1, and (-1)(-1)(-1) = -1 → (A).
Three negatives multiplied give a negative — Grade 7 rules for multiplying signed numbers.
7.NS.A.2Look For A PatternCheck with a=b=c=0: · · = = -1.
Plugging in 0s makes the arithmetic tiny and confirms the simplification.
7.NS.A.2Guess And Check(C)(D)(E) contain abc but our result is a constant, so they're out; the sign check picks (A) over (B).
If two expressions agree for one valid input, the one that depends on a, b, c must give a different value somewhere — but a constant doesn't.
6.EE.A.4Eliminate PossibilitiesThis AMC 10 problem only needs Grade 7 "multiplying signed numbers" you already know — each numerator is the opposite of one denominator, so each pair gives -1, and three of those multiply to -1.