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2020 AMC 10A Problem 3

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2020 AMC 10A Problem 3AMC 10 Step by Step
Name:                             Date: October 28, 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 top is a bottom with its sign flipped, like a - 3 and 3 - a. Matched up, each fraction is -1, so the product is -1.

Solution

The tops and bottoms look almost the same, just written backwards. So try to put each top over its matching bottom.

1. When you multiply fractions, you can swap the bottoms around freely: it is all one big product. So rewrite the expression as
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}.
2. Look at the first fraction. Try a number: if a=10a = 10, it is 7−7=−1\frac{7}{-7} = -1. In general 3−a3 - a is the negative of a−3a - 3, so the fraction is always −1-1. The problem says a≠3a \ne 3, so we never divide by zero.
3. The same thing happens with bb and with cc. Each fraction is −1-1.
4. Multiply: (−1)(−1)(−1)=−1(-1)(-1)(-1) = -1.

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

Why this works

Whenever you see p−qp - q and q−pq - p in the same expression, their ratio is −1-1. In a product you may re-pair tops and bottoms any way you like, so look for such pairs. Then count the minus signs: an odd number of them leaves the answer negative.

Alternative approach

Plug in easy numbers. Take 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.
Now check the choices with these numbers: (B) is 11, (C) is 12060=2\frac{120}{60} = 2, and (D) and (E) are ±(1120−160)\pm\left(\frac{1}{120} - \frac{1}{60}\right), which are small fractions. Only (A) gives −1-1.

The trap

Losing count of the minus signs. Three factors of -1 multiply to -1, not 1, so choice (B) is wrong.

Concise 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}.

Review in the AoPS books

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