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 "2021 AMC 12A Problems/Problem 4"

m (→‎Video Solution (Using Logic to Eliminate Choices))
m (→‎Video Solution by OmegaLearn(Using Logic to Eliminate Choices))
 
Line 125: Line 125:
 
https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
  
−
== Video Solution by OmegaLearn(Using Logic to Eliminate Choices) ==
+
== Video Solution by OmegaLearn (Using Logic to Eliminate Choices) ==
 
https://youtu.be/Mofw3VXHPyg
 
https://youtu.be/Mofw3VXHPyg
  

Latest revision as of 21:45, 28 October 2022

The following problem is from both the 2021 AMC 10A #7 and 2021 AMC 12A #4, so both problems redirect to this page.

Problem

Tom has a collection of $13$ snakes, $4$ of which are purple and $5$ of which are happy. He observes that

  • all of his happy snakes can add,
  • none of his purple snakes can subtract, and
  • all of his snakes that can't subtract also can't add.

Which of these conclusions can be drawn about Tom's snakes?

$\textbf{(A) }$ Purple snakes can add.

$\textbf{(B) }$ Purple snakes are happy.

$\textbf{(C) }$ Snakes that can add are purple.

$\textbf{(D) }$ Happy snakes are not purple.

$\textbf{(E) }$ Happy snakes can't subtract.

Solution 1 (Comprehensive Explanation of Logic)

We are given that \begin{align*} \text{happy}&\Longrightarrow\text{can add}, &(1) \\ \text{purple}&\Longrightarrow\text{cannot subtract}, \hspace{15mm} &(2) \\ \text{cannot subtract}&\Longrightarrow\text{cannot add}. &(3) \end{align*} Two solutions follow from here:

Solution 1.1 (Intuitive)

Combining $(2)$ and $(3)$ gives \begin{align*} \text{happy}&\Longrightarrow\text{can add}, &(1) \\ \lefteqn{\underbrace{\phantom{\text{purple}\Longrightarrow\text{cannot subtract}}}_{(2)}}\text{purple}&\Longrightarrow\overbrace{\text{cannot subtract}\Longrightarrow\text{cannot add}}^{(3)}. \hspace{2.5mm} &(*) \end{align*} Clearly, the answer is $\boxed{\textbf{(D)}}.$

~MRENTHUSIASM (credit given to abhinavg0627)

Solution 1.2 (Rigorous)

Recall that every conditional statement $\boldsymbol{p\Longrightarrow q}$ is always logically equivalent to its contrapositive $\boldsymbol{\lnot q\Longrightarrow\lnot p.}$

Combining $(1),(2)$ and $(3)$ gives \[\lefteqn{\underbrace{\phantom{\text{purple}\Longrightarrow\text{cannot subtract}}}_{(2)}}\text{purple}\Longrightarrow\overbrace{\text{cannot subtract}\Longrightarrow\lefteqn{\underbrace{\phantom{\text{cannot add}\Longrightarrow\text{not happy}}}_{\text{Contrapositive of }(1)}}\text{cannot add}}^{(3)}\Longrightarrow\text{not happy}. \hspace{15mm}(**)\] Applying the hypothetical syllogism to $(**),$ we conclude that \[\text{purple}\Longrightarrow\text{not happy},\] whose contrapositive is \[\text{happy}\Longrightarrow\text{not purple}.\] Therefore, the answer is $\boxed{\textbf{(D)}}.$

Remark

The conclusions in the other choices do not follow from $(**):$

$\textbf{(A) }\text{purple}\Longrightarrow\text{can add}$

$\textbf{(B) }\text{purple}\Longrightarrow\text{happy}$

$\textbf{(C) }\text{can add}\Longrightarrow\text{purple}$

$\textbf{(E) }\text{happy}\Longrightarrow\text{cannot subtract}$

~MRENTHUSIASM

Solution 2 (Process of Elimination)

From Solution 1.1, we can also see this through the process of elimination. Statement $A$ is false because purple snakes cannot add. $B$ is false as well because since happy snakes can add and purple snakes can not add, purple snakes are not happy snakes. $E$ is false using the same reasoning, purple snakes are not happy snakes so happy snakes can subtract since purple snakes cannot subtract. $C$ is false since snakes that can add are happy, not purple. That leaves statement D. $\boxed{\textbf{(D)}}$ is the only correct statement.

~Bakedpotato66

Solution 3 (Rigorous)

We first convert each statement to "If X, then Y" form:

  • If a snake is happy, then it can add.
  • If a snake is purple, then it can't subtract.
  • If a snake can't subtract, then it can't add.

Now, we simply check the truth value for each statement:

  1. Combining the last two propositions, we have
    • If a snake is purple, then it can't add.

    Thus, $\textbf{(A)}$ is never true.

  2. From the last part, we found that
    • If a snake is purple, then it can't add.

    Also, since the contrapositive of a proposition has the same truth value as the proposition itself, we know, from the first statement, that
    • If a snake can't add, then it isn't happy.

    Combining these two propositions, we find that
    • If a snake is purple, then it isn't happy. Purple snakes are not happy.

    Thus, $\textbf{(B)}$ is never true.

  3. From part $\textbf{(A)},$ we found that "If a snake is purple, then it can't add." This implies its contrapositive, "If a snake can add, then it is not purple." is true, meaning $\textbf{(C)}$ is NEVER true. [Thanks again to MRENTHUSIASM for pointing this out!]

  4. From the first statement, we have
    • If a snake is happy, then it can add.

    From the contrapositive of the third statement, we have
    • If a snake can add, then it can subtract.

    Then, from the contrapositive of the second statement, we have
    • If a snake can subtract, then it is not purple.

    Combining all of these yields
    • If a snake is happy, then it is not purple.

    Thus, $\textbf{(D)}$ is always true.

  5. From the first proposition, we have
    • If a snake is happy, then it can add.

    From the contrapositive of the third proposition, we have
    • If a snake can add, then it can subtract.

    Combining these two propositions gives
    • If a snake is happy, then it can subtract.

    Thus, $\textbf{(E)}$ is never true.

Therefore, $\boxed{\textbf{(D)}}$ is our answer.

~ Peace09 (My First Wiki Solution!)

~ MRENTHUSIASM (Revision Suggestions and Code Adjustments)

Video Solution (Simple & Quick)

https://youtu.be/hJKHaIcyIxA

~ Education the Study of Everything

Video Solution by Aaron He (Sets)

https://www.youtube.com/watch?v=xTGDKBthWsw&t=164

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=MUHja8TpKGw&t=259s (Note that there's a slight error in the video I corrected in the description)

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by OmegaLearn (Using Logic to Eliminate Choices)

https://youtu.be/Mofw3VXHPyg

~ pi_is_3.14

Video Solution

https://youtu.be/uDJv06-cNrI

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/s6E4E06XhPU?t=202 (AMC10A)

https://youtu.be/rEWS75W0Q54?t=353 (AMC12A)

~IceMatrix

Video Solution by The Learning Royal

https://youtu.be/AWjOeBFyeb4

See also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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
2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 12 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=2021_AMC_12A_Problems/Problem_4&oldid=179656"