2020 AMC 10A Problems/Problem 3

Problem

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

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

Solution 1

Note that $a-3$ is $-1$ times $3-a$. Likewise, $b-4$ is $-1$ times $4-b$ and $c-5$ is $-1$ times $5-c$. Therefore, the product of the given fraction equals $(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}$.

Solution 2

Substituting values for $a, b,\text{ and } c$, we see that if each of them satisfy the inequalities above, the value goes to be $-1$. Therefore, the product of the given fraction equals $(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}$.

Solution 3

It is known that $\frac{x-y}{y-x}=-1$ for $x\ne y$. We use this fact to cancel out the terms: $$\frac{\overset{-1}{\cancel{a-3}}}{\underset{1}{\xcancel{5-c}}} \cdot \frac{\overset{-1}{\bcancel{b-4}}}{\underset{1}{\cancel{3-a}}} \cdot \frac{\overset{-1}{\xcancel{c-5}}}{\underset{1}{\bcancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}.$$

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