Competition · AMC preparation · step 4 of 4
AMC 10 · 2020A · #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: (a-3)/(3-a) · (b-4)/(4-b) · (c-5)/(5-c). 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.
Spot the opposite pairs
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.
Reversing the order in a subtraction flips its sign.
▸ Why?
Adding two numbers in either order gives the same result, so only the subtraction feels the swap.
▸ Why?
The two results are opposites, so pairing them off leaves only a sign to carry.
Pair each numerator with its match
Reorder 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 WaysMultiply the three negative ones
Each 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 PatternTest with real numbers
Check with a=b=c=0: · · = = -1.
Plugging in 0s makes the arithmetic tiny and confirms the simplification.
7.NS.A.2Guess And CheckRule out the variable choices
(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.
- Spot the opposite pairs
- Pair each numerator with its match
- Multiply the three negative ones
- Test with real numbers
- Rule out the variable choices
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