2022 AMC 10A Problem 12

P11-15~5 minPrint

Logic Puzzles Systems of Equations Split into exhaustive cases and handle each Start from the end state / desired conclusion and reverse

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?

Show solution without answering

Key insight

Write down how each kind of child answers the three questions. Truth-tellers say yes, no, no; liars say yes, yes, no; alternaters say no, no, no or yes, yes, yes.

Solution

Step 1: how does each kind of child answer? The true answers to "Are you a truth-teller / an alternater / a liar?" depend on the child, and then we apply lying or not.

A truth-teller answers honestly: yes, no, no.
A liar's honest answers would be no, no, yes; a liar says the opposite: yes, yes, no.
An alternater who starts with the truth: question 1 honestly "no"; question 2 is a lie, so "no" (the truth is yes); question 3 honestly "no". So: no, no, no.
An alternater who starts with a lie: question 1 "yes" (a lie); question 2 honestly "yes"; question 3 "yes" (a lie). So: yes, yes, yes.

Kind of child	Q1	Q2	Q3
Truth-teller	yes	no	no
Liar	yes	yes	no
Alternater, truth first	no	no	no
Alternater, lie first	yes	yes	yes

Step 2: count, starting with the question the fewest kinds say yes to. That is question 3.

Question 3: only lie-first alternaters say yes, so there are $9$ of them.
Question 2: liars and lie-first alternaters say yes. So the number of liars is $15 - 9 = 6$ .
Question 1: truth-tellers, liars and lie-first alternaters say yes. So the number of truth-tellers is $22 - 6 - 9 = 7$ .
Check: the truth-first alternaters number $31 - 7 - 6 - 9 = 9$ , a sensible count.

Step 3: candy. A truth-teller says yes only to question 1, so each gets exactly one piece. The $7$ truth-tellers get $7$ pieces in all.

The answer is $\boxed{\textbf{(A)}\ 7}$ .

Why this works

Truth-teller and liar puzzles turn into simple counting once you make a table of how each kind answers. The alternaters split into two kinds depending on how they start. Then begin with the question that the fewest kinds answer yes to, and work back up one question at a time.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Work out how each kind of child answers the three questions (truth-teller? alternater? liar?).

Truth-teller: yes, no, no.
Liar: the true answers are no, no, yes; a liar negates them: yes, yes, no.
Alternater starting truthful: truthfully "no" to question 1, then lies about being an alternater: "no", then truthfully "no": no, no, no.
Alternater starting with a lie: yes, yes, yes.

Let $T$ , $L$ , $A_1$ , $A_2$ be the numbers of truth-tellers, liars, truth-first alternaters, and lie-first alternaters. Reading the counts from the last question backward:

Question 3 ( $9$ yeses): only lie-first alternaters said yes, so $A_2 = 9$ .
Question 2 ( $15$ yeses): liars and lie-first alternaters, so $L + 9 = 15$ , $L = 6$ .
Question 1 ( $22$ yeses): truth-tellers, liars, lie-first alternaters, so $T + 6 + 9 = 22$ , $T = 7$ .

(Then $A_1 = 31 - 22 = 9$ , consistent.) Each truth-teller received exactly one piece of candy, on question 1, so the truth-tellers received $7$ pieces in all.

The answer is $\boxed{\textbf{(A)}\ 7}$ .

Review in the AoPS books

Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.4)
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.4)