Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3: Made the solution COLORFUL)
(→‎Solution 3 (Fastest))
 
(28 intermediate revisions by 5 users not shown)
Line 3: Line 3:
 
<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>
+
<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 ==
+
== Solution 1 (Negatives) ==
−
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>.
+
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>
 +
~CoolJupiter ~MRENTHUSIASM
  
−
== Solution 2 ==
+
== Solution 2 (Answer Choices) ==
−
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>.
+
At <math>(a,b,c)=(4,5,6),</math> the answer choices become
−
Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>.
 
  
−
== Solution 3 ==
+
<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>
−
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>
 
−
~CoolJupiter (Solution)
 
  
−
~MRENTHUSIASM (<math>\LaTeX</math> Adjustments)
+
and the original expression becomes <cmath>\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{\textbf{(A) } {-}1}.</cmath>
 +
~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

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 (Negatives)

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 ~MRENTHUSIASM

Solution 2 (Answer Choices)

At $(a,b,c)=(4,5,6),$ the answer choices become

$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } {-}\frac{1}{120} \qquad \textbf{(E) } \frac{1}{120}$

and the original expression becomes \[\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{\textbf{(A) } {-}1}.\] ~MRENTHUSIASM

Solution 3 (Fastest)

We can simply set $x = a - 3, y = b - 4,$ and $z = 5 - c$. Now, the problem simplifies to \[\frac{x}{z}\cdot\frac{-y}{x}\cdot\frac{z}{y}=\frac{-xyz}{xyz}=\boxed{\textbf{(A) } {-}1}.\]

Explanation: After substituting $x$, $y$, and $z$, the opposites (for example $5 - c$ and $c - 5$) can just be written as the negative of each. With the same example, this can be shown by: $c - 5 = -(5 - c)$.

~GREATEST

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
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_3&oldid=233205"