AMC 10 · 2020 · #3

Grade 7 algebra

Archetype Arithmetic Expression Decomposition

fraction-arithmeticpattern-recognitionpolynomial-factoring pattern-recognition ↑ Prerequisites: fraction-arithmetic

📏 Short solution 💡 2 insights

Problem

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

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

Pick an answer.

(A)

${-}1$

(B)

(C)

$\frac{abc}{60}$

(D)

$\frac{1}{abc} - \frac{1}{60}$

(E)

$\frac{1}{60} - \frac{1}{abc}$

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Toolkit + CCSS Solution

Understand

Restated: Simplify the product $\dfrac{a-3}{5-c} \cdot \dfrac{b-4}{3-a} \cdot \dfrac{c-5}{4-b}$, given that none of the denominators are zero.

Givens: Three fractions multiplied together; Numerators: $a-3$, $b-4$, $c-5$; Denominators: $5-c$, $3-a$, $4-b$; $a \neq 3$, $b \neq 4$, $c \neq 5$ (no division by zero); Answer choices: (A) $-1$, (B) $1$, (C) $\frac{abc}{60}$, (D) $\frac{1}{abc} - \frac{1}{60}$, (E) $\frac{1}{60} - \frac{1}{abc}$

Unknowns: The simplest form of the product

Understand

Restated: Simplify the product $\dfrac{a-3}{5-c} \cdot \dfrac{b-4}{3-a} \cdot \dfrac{c-5}{4-b}$, given that none of the denominators are zero.

Plan

Primary tool: #5 Look for a Pattern

Secondary: #15 Reorganize Information, #6 Guess and Check, #3 Eliminate Possibilities

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: $\frac{a-3}{3-a} \cdot \frac{b-4}{4-b} \cdot \frac{c-5}{5-c}$. 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.

Execute — Answer: A

#5 Look for a Pattern 6.NS.C.5 Step 1

Pattern-spot: each numerator is the negative of one of the denominators.
$(a - 3) = -(3 - a)$, $(b - 4) = -(4 - b)$, $(c - 5) = -(5 - c)$.

$$a - 3 = -(3 - a), \quad b - 4 = -(4 - b), \quad c - 5 = -(5 - c)$$

💡 Reversing the order in a subtraction flips its sign — Grade 6 understanding of opposite (negative) numbers.

#15 Reorganize Information 3.OA.B.5 Step 2

Reorganize the multiplication so each numerator sits over its own opposite.
Because multiplication can be reordered, pair $(a-3)$ with $(3-a)$, $(b-4)$ with $(4-b)$, $(c-5)$ with $(5-c)$.

$$\dfrac{a-3}{5-c} \cdot \dfrac{b-4}{3-a} \cdot \dfrac{c-5}{4-b} = \dfrac{a-3}{3-a} \cdot \dfrac{b-4}{4-b} \cdot \dfrac{c-5}{5-c}$$

💡 Commutative property of multiplication — change the pairing to expose the pattern.

#5 Look for a Pattern 7.NS.A.2 Step 3

Each repaired fraction is $\frac{\text{thing}}{-\text{same thing}} = -1$.
Multiply: $(-1)(-1)(-1) = -1$.

$$(-1)(-1)(-1) = -1 \;\Rightarrow\; \textbf{(A)}$$

💡 Three negatives multiplied give a negative — Grade 7 rules for multiplying signed numbers.

#6 Guess and Check 7.NS.A.2 Step 4

Confirm with concrete numbers.
Pick $a = 0, b = 0, c = 0$ (all valid since $0 \neq 3, 4, 5$).
The product becomes $\frac{-3}{5} \cdot \frac{-4}{3} \cdot \frac{-5}{4} = \frac{(-3)(-4)(-5)}{5 \cdot 3 \cdot 4} = \frac{-60}{60} = -1$.

$$\dfrac{-3}{5} \cdot \dfrac{-4}{3} \cdot \dfrac{-5}{4} = \dfrac{-60}{60} = -1 \;\Rightarrow\; \textbf{(A)}$$

💡 Plugging in $0$s makes the arithmetic tiny and confirms the simplification.

#3 Eliminate Possibilities 6.EE.A.4 Step 5

Eliminate the variable choices.
(C), (D), (E) all contain $abc$, but our simplification gave a constant $-1$ for all valid inputs.
So those can't be right.
Between (A) $-1$ and (B) $+1$, our work shows $-1$.

$$\text{constant result} \Rightarrow \text{ choice is (A) or (B); sign check picks } \textbf{(A)}$$

💡 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.

[1] #5 6.NS.C.5 Pattern-spot: each numerator is the negative of one of the denominators. $(a - 3

[2] #15 3.OA.B.5 Reorganize the multiplication so each numerator sits over its own opposite. Beca

[3] #5 7.NS.A.2 Each repaired fraction is $\frac{\text{thing}}{-\text{same thing}} = -1$. Multip

[4] #6 7.NS.A.2 Confirm with concrete numbers. Pick $a = 0, b = 0, c = 0$ (all valid since $0 \n

[5] #3 6.EE.A.4 Eliminate the variable choices. (C), (D), (E) all contain $abc$, but our simplif

Review

Reasonableness: Try a second concrete triple to be sure. With $a = 1, b = 1, c = 1$: product $= \frac{-2}{4} \cdot \frac{-3}{2} \cdot \frac{-4}{3} = \frac{(-2)(-3)(-4)}{4 \cdot 2 \cdot 3} = \frac{-24}{24} = -1$. Same answer, supporting (A).

Alternative: Tool #6 (Guess and Check) alone, as in the previous step, would have settled the problem in under a minute. For AMC multiple-choice, plugging in convenient values is often faster than algebraic restructuring.

CCSS standards used (min grade 7)

3.OA.B.5 Apply properties of operations as strategies to multiply and divide (Using the commutative property to repair-pair the three fractions $\frac{a-3}{3-a}, \frac{b-4}{4-b}, \frac{c-5}{5-c}$.)
6.NS.C.5 Understand that positive and negative numbers describe quantities (Recognizing that $(a-3)$ and $(3-a)$ are opposites — reversing subtraction flips the sign.)
6.EE.A.4 Identify when two expressions are equivalent (Using one concrete input to rule out the choices that still depend on $a, b, c$.)
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers (Computing $(-1)(-1)(-1) = -1$ and verifying with signed fractions.)

⭐ 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 $-1$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 7)

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