Difference between revisions of "2020 AMC 10A Problems/Problem 3"
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<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> | ||
~CoolJupiter (Solution) | ~CoolJupiter (Solution) | ||
+ | |||
+ | ~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 == | ||
+ | At <math>(a,b,c)=(4,5,6),</math> the answer choices become | ||
+ | |||
+ | <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> | ||
+ | ~MRENTHUSIASM | ||
== Video Solution 1 == | == Video Solution 1 == |
Revision as of 18:23, 4 November 2021
Contents
Problem
Assuming , , and , what is the value in simplest form of the following expression?
Solution 1
If then We use this fact to simplify the original expression: ~CoolJupiter (Solution)
~MRENTHUSIASM ( Adjustments)
Solution 2
Note that is times Likewise, is times and is times Therefore, the product of the given fraction equals
Solution 3
At the answer choices become
and the original expression becomes ~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.