AMC 10 · 2020 · #3

Grade 7 algebra
fraction-arithmeticpattern-recognitionpolynomial-factoring pattern-recognition ↑ Prerequisites: fraction-arithmetic
📏 Short solution 💡 2 insights
Problem
Simplify the product a35c\frac{a-3}{5-c} · b43a\frac{b-4}{3-a} · c54b\frac{c-5}{4-b}, given that none of the denominators are zero.

Pick an answer.

(A)
${-}1$
(B)
1
(C)
$\frac{abc}{60}$
(D)
$\frac{1}{abc} - \frac{1}{60}$
(E)
$\frac{1}{60} - \frac{1}{abc}$

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

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: a33a\frac{a-3}{3-a} · b44b\frac{b-4}{4-b} · c55c\frac{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.

1STEP 1

Each numerator is the opposite of a denominator: (a-3) = -(3-a), same for the others.

a - 3 = -(3 - a), b - 4 = -(4 - b), c - 5 = -(5 - c)
2STEP 2

Reorder the factors so each numerator sits over its own opposite denominator.

a35c\frac{a-3}{5-c} · b43a\frac{b-4}{3-a} · c54b\frac{c-5}{4-b} = a33a\frac{a-3}{3-a} · b44b\frac{b-4}{4-b} · c55c\frac{c-5}{5-c}
3STEP 3

Each re-paired fraction is thingthing\frac{thing}{-thing} = -1, and (-1)(-1)(-1) = -1 → (A).

(-1)(-1)(-1) = -1 → (A)
4STEP 4

Check with a=b=c=0: 35\frac{-3}{5} · 43\frac{-4}{3} · 54\frac{-5}{4} = 6060\frac{-60}{60} = -1.

35\frac{-3}{5} · 43\frac{-4}{3} · 54\frac{-5}{4} = 6060\frac{-60}{60} = -1 → (A)
5STEP 5

(C)(D)(E) contain abc but our result is a constant, so they're out; the sign check picks (A) over (B).

constant result → choice is (A) or (B); sign check picks (A)
Answer
-1
Try a second concrete triple to be sure. With a = 1, b = 1, c = 1: product = 24\frac{-2}{4} · 32\frac{-3}{2} · 43\frac{-4}{3} = (2)(3)(4)(423)\frac{(-2)(-3)(-4)}{(4 · 2 · 3)} = 2424\frac{-24}{24} = -1. Same answer, supporting (A).
💡Key takeaway

This 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.