# 2021 AMC 12A Problems/Problem 4

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)

~ Education the Study of Everything

## 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)

~ pi_is_3.14

~savannahsolver

~IceMatrix

## Video Solution by The Learning Royal

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 