Difference between revisions of "2020 AMC 10A Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | Assuming <math>a\neq3</math>, <math>b\neq4</math>, and <math>c\neq5</math>, what is the value in simplest form of the following expression | + | Assuming <math>a\neq3</math>, <math>b\neq4</math>, and <math>c\neq5</math>, what is the value in simplest form of the following expression? |
<cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath> | <cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath> | ||
| − | + | <math>\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}</math> | |
== Solution 1 (Negatives) == | == Solution 1 (Negatives) == | ||
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> |
~CoolJupiter ~MRENTHUSIASM | ~CoolJupiter ~MRENTHUSIASM | ||
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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) } -1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } -\frac{1}{120} \qquad \textbf{(E) } \frac{1}{120}</math> | + | <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> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | == Solution 3 (Fastest) == | ||
| + | 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}=\frac{-xyz}{xyz}=\boxed{\textbf{(A) } {-}1}.</cmath> | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | ~GREATEST | ||
== Video Solution 1 == | == Video Solution 1 == | ||
Latest revision as of 01:04, 11 November 2024
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
Solution 3 (Fastest)
We can simply set 
and 
. Now, the problem simplifies to ![\[\frac{x}{z}\cdot\frac{-y}{x}\cdot\frac{z}{y}=\frac{-xyz}{xyz}=\boxed{\textbf{(A) } {-}1}.\]](http://latex.artofproblemsolving.com/9/2/2/9227cd7bd2e88f44dfbad50c6b21a871793a8cd3.png)
Explanation: After substituting 
, 
, and 
, the opposites (for example 
and 
) can just be written as the negative of each. With the same example, this can be shown by: 
.
~GREATEST
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. 
