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

m (→‎Solution 2)
(Ordered the video solutions. Also, I fixed the coding in Solution 3 so that the -1's are on top of the factors instead of appearing in the same row as the numerator. Relatively, this version is clearer.)
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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>.
+
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 ==
 
== Solution 2 ==
 
Substituting values for <math>a, b,\text{ and } c</math>, we see that if each of them satisfy the inequalities above, the value goes to be <math>-1</math>.  
 
Substituting values for <math>a, b,\text{ and } 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>.
+
Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>.
  
 
== Solution 3 ==
 
== Solution 3 ==
−
It is known that <math>\frac{x-y}{y-x}=-1</math> for <math>x\ne y</math>. We use this fact to cancel out the terms.
+
It is known that <math>\frac{x-y}{y-x}=-1</math> for <math>x\ne y</math>. We use this fact to cancel out the terms: <cmath>\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}.</cmath>
  
−
<math>\frac{\cancel{(a-3)} -1 \cancel{(b-4)} -1 \cancel{(c-5)} -1}{\cancel{(5-c)}\cancel{(3-a)}\cancel{(4-b)}}=(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}</math>  
+
~CoolJupiter (Solution)
−
 
+
~MRENTHUSIASM (<math>\LaTeX</math> Adjustment)
−
~CoolJupiter
 
−
 
 
−
== Video Solution 5==
 
−
https://youtu.be/ba6w1OhXqOQ?t=956
 
−
 
 
−
~ pi_is_3.14
 
  
 
== Video Solution 1 ==
 
== Video Solution 1 ==
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~savannahsolver
 
~savannahsolver
 +
 +
== Video Solution 5==
 +
https://youtu.be/ba6w1OhXqOQ?t=956
 +
 +
~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2020|ab=A|num-b=2|num-a=4}}
 
{{AMC10 box|year=2020|ab=A|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:16, 17 April 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

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: \[\frac{\overset{-1}{\cancel{a-3}}}{\underset{1}{\xcancel{5-c}}} \cdot \frac{\overset{-1}{\bcancel{b-4}}}{\underset{1}{\cancel{3-a}}} \cdot \frac{\overset{-1}{\xcancel{c-5}}}{\underset{1}{\bcancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}.\]

~CoolJupiter (Solution) ~MRENTHUSIASM ($\LaTeX$ Adjustment)

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)
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Problem 2 		Followed by
Problem 4
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All AMC 10 Problems and Solutions

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