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

(→‎Solution 3: Made the solution COLORFUL)
(Ordered the solutions by elegance. PM me if you disagree with this order--I pushed the observation solutions to the end.)
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== Solution 1 ==
 
== Solution 1 ==
−
Note that <math>a-3</math> is <math>-1</math> times <math>3-a</math>. Likewise, <math>b-4</math> is <math>-1</math> times <math>4-b</math> and <math>c-5</math> is <math>-1</math> times <math>5-c</math>. Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>.
 
−
 
−
== Solution 2 ==
 
−
Substituting values for <math>a, b,</math> and <math>c</math>, we see that if each of them satisfy the inequalities above, the value goes to be <math>-1</math>.
 
−
Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>.
 
−
 
−
== Solution 3 ==
 
 
If <math>x\neq y,</math> then <math>\frac{x-y}{y-x}=1.</math> We use this fact to simplify the original expression:
 
If <math>x\neq y,</math> then <math>\frac{x-y}{y-x}=1.</math> We use this fact to simplify the original expression:
 
<cmath>\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}.</cmath>
 
<cmath>\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}.</cmath>
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~MRENTHUSIASM (<math>\LaTeX</math> Adjustments)
 
~MRENTHUSIASM (<math>\LaTeX</math> Adjustments)
 +
 +
== Solution 2 ==
 +
Note that <math>a-3</math> is <math>-1</math> times <math>3-a</math>. Likewise, <math>b-4</math> is <math>-1</math> times <math>4-b</math> and <math>c-5</math> is <math>-1</math> times <math>5-c</math>. Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>.
 +
 +
== Solution 3 ==
 +
Substituting values for <math>a, b,</math> and <math>c</math>, we see that if each of them satisfy the inequalities above, the value goes to be <math>-1</math>.
 +
Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>.
  
 
== Video Solution 1 ==
 
== Video Solution 1 ==

Revision as of 15:39, 27 August 2021

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

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 (Solution)

~MRENTHUSIASM ($\LaTeX$ Adjustments)

Solution 2

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 3

Substituting values for $a, b,$ 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 $(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}$.

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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