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2020 AMC 10A Problem 3

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2020 AMC 10A Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Algebraic Manipulation · difficulty 1 of 5 · about 1 min · position P1-10

Assuming a≠3a\neq3 , b≠4b\neq4 , and c≠5c\neq5 , what is the value in simplest form of the following expression? a−35−c⋅b−43−a⋅c−54−b\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}

  1. A)

    −1{-}1

  2. B)

    11

  3. C)

    abc60\frac{abc}{60}

  4. D)

    1abc−160\frac{1}{abc} - \frac{1}{60}

  5. E)

    160−1abc\frac{1}{60} - \frac{1}{abc}

Answer

A

Key insight

Each numerator is the negative of some denominator: (a-3)/(3-a) = -1, and likewise for b and c, so the product is (-1)^3.

Solution

The product of three fractions does not care how the numerators and denominators are paired, so regroup them so that matching expressions sit together:
a−33−a⋅b−44−b⋅c−55−c.\frac{a-3}{3-a} \cdot \frac{b-4}{4-b} \cdot \frac{c-5}{5-c}.
Because 3−a=−(a−3)3 - a = -(a - 3), the first fraction equals −1-1; the hypotheses a≠3a \ne 3 etc. guarantee no denominator is zero. The same holds for the other two. The product is
(−1)(−1)(−1)=−1.(-1)(-1)(-1) = -1.

The answer is (A) −1\boxed{\textbf{(A)}\ -1}.

Why this works

Whenever an expression contains both p−qp - q and q−pq - p, they cancel to −1-1. Since multiplication is commutative, you are free to re-pair numerators with denominators to expose these cancellations. Count the sign flips carefully: an odd number of them leaves a negative.

Alternative approach

Plug in convenient values such as a=4a = 4, b=5b = 5, c=6c = 6: the expression becomes 1−1⋅1−1⋅1−1=−1\frac{1}{-1}\cdot\frac{1}{-1}\cdot\frac{1}{-1} = -1. Only choice (A) matches (choices (C), (D), (E) would give 22, 1120−160\frac{1}{120}-\frac{1}{60}, and its negative).

The trap

Losing track of the number of sign flips and answering 1 (choice B).

Common mistakes

  • Losing track of the number of sign flips and answering 1 (choice B).
  • Multiplying the numerators and denominators out in full and trying to simplify a cubic over a cubic.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Test small/specific values or special cases to find or verify the answer