# 2022 AMC 10A Problems/Problem 12

## Problem

On Halloween $31$ children walked into the principal's office asking for candy. They can be classified into three types: Some always lie; some always tell the truth; and some alternately lie and tell the truth. The alternaters arbitrarily choose their first response, either a lie or the truth, but each subsequent statement has the opposite truth value from its predecessor. The principal asked everyone the same three questions in this order.

"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.

"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.

"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who answered yes.

How many pieces of candy in all did the principal give to the children who always tell the truth? $\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$

## Solution 1

Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all $9$ children that answered yes are alternaters that falsely answer Questions 1 and 3, and truthfully answer Question 2. The rest of the alternaters, however many there are, have the opposite behavior.

Consider the second question, "Are you an alternater?": The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that $9$ alternaters truthfully answer here. Because only liars and $9$ alternaters answer yes, we can deduce that there are $15-9=6$ liars.

Consider the first question, "Are you a truth teller?": Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that $9$ alternaters lie here. From the previous part, we know that there are $6$ liars. Because only the number of truth tellers is unknown here, we can deduce that there are $22-9-6=7$ truth tellers.

The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers only answer yes on the first question, we know that all $7$ of them said yes once, resulting in $\boxed{\textbf{(A) } 7}$ pieces of candy.

~phuang1024

## Solution 2

On the first question, truth tellers would say yes. Liars would say yes because they are not truth tellers and thus will say the opposite of no. Some alternaters may lie on this question too, meaning they say yes.

On the second question, Liars would say yes because they are not alternaters and thus will say the opposite of no. Alternaters only say yes to this question if and only if they said yes to the previous question. Thus, the difference between the amount of people that said yes first and that said yes second is the amount of truth tellers, which is $22-15=7$. Because it is obvious that truth tellers only say yes on the first question, our answer is $\boxed{\textbf{(A) } 7}$.

~sigma

## Solution 3

Suppose that there are $T$ truth-tellers, $L$ liars, $A$ alternaters who answered yes-no-yes, and $A'$ alternaters who answered no-yes-no.

We have the following system of equations: $$\begin{array}{ccccccccrcccc} T &+ &L &+ &A &+ &A' &= &31, & & & & (1) \\ [0.5ex] T &+ &L &+ &A & & &= &22, & & & & (2) \\ [0.5ex] & &L & & &+ &A' &= &15, & & & & (3) \\ [0.5ex] & & & & A & & &= &9. & & & & (4) \end{array}$$ Subtracting $(2)$ from $(1)$ gives $A'=9.$ From $(3),$ it follows that $L=15-A'=6.$ Finally, from $(2),$ we have $T=22-A-L=\boxed{\textbf{(A) } 7}.$

## Remark (Fake Solve)

This problem is broken in an interesting way, that helps the test-taker. Since $A=A'=9$, you can still get the correct answer if you misinterpret the problem as "alternaters alternate their answer (not their truth value)" or even "alternaters are arbitraters of two types: $A$ who answer arbitrarily but all give the same answer as each other, and $A'$ who all answer the opposite of $A$.

It's also notable that the misinterpretation makes the solution a bit less trivial, so that the solution actually relies on all the information. This suggests that the question-writer may have been mistaken but got lucky.

~oinava, based on demonstration of misinterpretation fakesolve by ~orenbad

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