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Difference between revisions of "2021 AMC 12A Problems/Problem 4"

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{{duplicate|[[2021 AMC 10A Problems#Problem 7|2021 AMC 10A #7]] and [[2021 AMC 12A Problems#Problem 4|2021 AMC 12A #4]]}}
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{{duplicate|[[2021 AMC 10A Problems/Problem 7|2021 AMC 10A #7]] and [[2021 AMC 12A Problems/Problem 4|2021 AMC 12A #4]]}}
  
 
==Problem==
 
==Problem==
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<math>\textbf{(E) }</math> Happy snakes can't subtract.
 
<math>\textbf{(E) }</math> Happy snakes can't subtract.
  
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==Solution 1==
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==Solution 1 (Comprehensive Explanation of Logic)==
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We know that purple snakes cannot subtract, thus they cannot add either. Since happy snakes must be able to add, the purple snakes cannot be happy. Therefore, we know that the happy snakes are not purple and the answer is <math>\boxed{\textbf{(D)}}</math>.
 
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--abhinavg0627
 
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==Solution 2 (Comprehensive Explanation Using Arrows)==
 
 
We are given that
 
We are given that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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Two solutions follow from here:
 
Two solutions follow from here:
  
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===Solution 2.1 (Intuitive)===
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===Solution 1.1 (Intuitive)===
 
Combining <math>(2)</math> and <math>(3)</math> gives
 
Combining <math>(2)</math> and <math>(3)</math> gives
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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Clearly, the answer is <math>\boxed{\textbf{(D)}}.</math>
 
Clearly, the answer is <math>\boxed{\textbf{(D)}}.</math>
  
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~MRENTHUSIASM
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~MRENTHUSIASM (credit given to abhinavg0627)
  
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===Solution 2.2 (Rigorous)===
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===Solution 1.2 (Rigorous)===
 
<i><b>Recall that every conditional statement <math>\boldsymbol{p\Longrightarrow q}</math> is always logically equivalent to its contrapositive <math>\boldsymbol{\lnot q\Longrightarrow\lnot p.}</math></b></i>
 
<i><b>Recall that every conditional statement <math>\boldsymbol{p\Longrightarrow q}</math> is always logically equivalent to its contrapositive <math>\boldsymbol{\lnot q\Longrightarrow\lnot p.}</math></b></i>
  
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 3 (Process of Elimination)==
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==Solution 2 (Process of Elimination)==
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From Solution 2.1, we can also see this through the process of elimination.  
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From Solution 1.1, we can also see this through the process of elimination.  
 
Statement <math>A</math> is false because purple snakes cannot add. <math>B</math> is false as well  because since happy snakes can add and purple snakes can not add, purple snakes are not happy snakes. <math>E</math> is false using the same reasoning, purple snakes are not happy snakes so happy snakes can subtract since purple snakes cannot subtract. <math>C</math> is false since snakes that can add are happy, not purple. That leaves statement D. <math>\boxed{\textbf{(D)}}</math> is the only correct statement.
 
Statement <math>A</math> is false because purple snakes cannot add. <math>B</math> is false as well  because since happy snakes can add and purple snakes can not add, purple snakes are not happy snakes. <math>E</math> is false using the same reasoning, purple snakes are not happy snakes so happy snakes can subtract since purple snakes cannot subtract. <math>C</math> is false since snakes that can add are happy, not purple. That leaves statement D. <math>\boxed{\textbf{(D)}}</math> is the only correct statement.
  
 
~Bakedpotato66
 
~Bakedpotato66
  
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==Solution 4 (Rigorous)==
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==Solution 3 (Rigorous)==
 
We first convert each statement to "If X, then Y" form:
 
We first convert each statement to "If X, then Y" form:
  
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https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
  
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== Video Solution (Using Logic to Eliminate Choices) ==
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== Video Solution by OmegaLearn (Using Logic to Eliminate Choices) ==
 
https://youtu.be/Mofw3VXHPyg
 
https://youtu.be/Mofw3VXHPyg
  
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~IceMatrix
 
~IceMatrix
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==Video Solution by The Learning Royal==
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https://youtu.be/AWjOeBFyeb4
  
 
==See also==
 
==See also==

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

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