Difference between revisions of "2020 AMC 10A Problems/Problem 3"
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At <math>(a,b,c)=(4,5,6),</math> the answer choices become | At <math>(a,b,c)=(4,5,6),</math> the answer choices become | ||
| − | <math>\textbf{(A) } | + | <math>\textbf{(A) } -1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } -\frac{1}{120} \qquad \textbf{(E) } \frac{1}{120}</math> |
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> | ||
Revision as of 13:55, 29 December 2022
Contents
Problem
Assuming 
, 
, and 
, what is the value in simplest form of the following expression?
![\[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]](http://latex.artofproblemsolving.com/9/2/6/926ff7e4cb0ce3392334f8215dd3aa4c51438f15.png)
Solution 1 (Negatives)
If 
then 
We use this fact to simplify the original expression:
![\[\frac{\color{red}\overset{-1}{\cancel{a-3}}}{\color{blue}\underset{1}{\cancel{5-c}}} \cdot \frac{\color{green}\overset{-1}{\cancel{b-4}}}{\color{red}\underset{1}{\cancel{3-a}}} \cdot \frac{\color{blue}\overset{-1}{\cancel{c-5}}}{\color{green}\underset{1}{\cancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } {-}1}.\]](http://latex.artofproblemsolving.com/6/7/7/6772947b7c8cd9af853e09759abd7ebc2173f282.png)
~CoolJupiter ~MRENTHUSIASM
Solution 2 (Answer Choices)
At 
the answer choices become
and the original expression becomes ![\[\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{\textbf{(A) } {-}1}.\]](http://latex.artofproblemsolving.com/6/9/b/69ba588f0dec9a22afadecee1e319cac8622cfc2.png)
~MRENTHUSIASM
Video Solution 1
~IceMatrix
Video Solution 2
Education, The Study of Everything
Video Solution 3
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
Video Solution 4
~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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
