# 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 (Negatives)

If $x\neq y,$ then $\frac{x-y}{y-x}=-1.$ We use this fact to simplify the original expression: $$\frac{\color{red}\overset{-1}{\cancel{a-3}}}{\color{blue}\underset{1}{\cancel{5-c}}} \cdot \frac{\color{green}\overset{-1}{\cancel{b-4}}}{\color{red}\underset{1}{\cancel{3-a}}} \cdot \frac{\color{blue}\overset{-1}{\cancel{c-5}}}{\color{green}\underset{1}{\cancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } {-}1}.$$ ~CoolJupiter ~MRENTHUSIASM

At $(a,b,c)=(4,5,6),$ the answer choices become $\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } {-}\frac{1}{120} \qquad \textbf{(E) } \frac{1}{120}$

and the original expression becomes $$\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{\textbf{(A) } {-}1}.$$ ~MRENTHUSIASM

~IceMatrix

## Video Solution 2

Education, The Study of Everything

~bobthefam

~savannahsolver

## Video Solution 5

~ pi_is_3.14

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 