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Difference between revisions of "2020 AMC 10A Problems/Problem 3"

(Sol 1 already says something very similar to Sol 3, and Sol 1 used color.)
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and the original expression becomes <cmath>\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{\textbf{(A) } {-}1}.</cmath>
 
and the original expression becomes <cmath>\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{\textbf{(A) } {-}1}.</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
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 +
== Solution 3 (Substitution) ==
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We can simply set <math>x = a - 3, y = b - 4,</math> and <math>z = 5 - c</math>. Now, the problem simplifies to <cmath>\frac{x}{z}\cdot\frac{-y}{x}\cdot\frac{z}{y}=\boxed{\textbf{(A) } {-}1}.</cmath>
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 +
 +
Explanation: After substituting <math>x</math>, <math>y</math>, and <math>z</math>, the opposites (for example <math>5 - c</math> and <math>c - 5</math>) can just be written as the negative of each. With the same example, this can be shown by: <math>c - 5 = -(5 - c)</math>.
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~GREATEST
  
 
== Video Solution 1 ==
 
== Video Solution 1 ==

Revision as of 23:38, 10 November 2024

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

Solution 2 (Answer Choices)

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

Solution 3 (Substitution)

We can simply set $x = a - 3, y = b - 4,$ and $z = 5 - c$. Now, the problem simplifies to \[\frac{x}{z}\cdot\frac{-y}{x}\cdot\frac{z}{y}=\boxed{\textbf{(A) } {-}1}.\]


Explanation: After substituting $x$, $y$, and $z$, the opposites (for example $5 - c$ and $c - 5$) can just be written as the negative of each. With the same example, this can be shown by: $c - 5 = -(5 - c)$.

~GREATEST

Video Solution 1

https://youtu.be/WUcbVNy2uv0

~IceMatrix

Video Solution 2

https://youtu.be/Nrdxe4UAqkA

Education, The Study of Everything

Video Solution 3

https://www.youtube.com/watch?v=7-3sl1pSojc

~bobthefam

Video Solution 4

https://youtu.be/ZccL6yKrTiU

~savannahsolver

Video Solution 5

https://youtu.be/ba6w1OhXqOQ?t=956

~ pi_is_3.14

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AMC 10 Problems and Solutions

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