AMC 10 · 2022 · #12

Grade 6 arithmetic

logical-deductionsystems-of-equationscasework caseworkconvert-to-algebra ↑ Prerequisites: logical-deduction

📏 Medium solution 💡 3 insights

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?

$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: $31$ children come in three flavors — truth-tellers (always say true things), liars (always say false things), and alternaters (flip between truth and lie, starting with either). The principal asks each child the same three Yes/No questions in order: (1) Are you a truth-teller? (2) Are you an alternater? (3) Are you a liar? After each question, every child who said Yes gets one candy. The three Yes-counts are $22$, $15$, $9$. How much total candy went to the truth-tellers?

Givens: Total children: $31$; Yes-counts: question 1 -> $22$, question 2 -> $15$, question 3 -> $9$; Truth-teller always tells the truth; liar always lies; alternater alternates and may start either way

Unknowns: Total candies received by the truth-tellers

Understand

Plan

Primary tool: #4 Use Matrix Logic

Secondary: #2 Make a Systematic List, #7 Identify Subproblems, #13 Convert to Algebra, #3 Eliminate Possibilities

Tool #4 (Matrix Logic) is the natural fit: build a small grid whose rows are the four kinds of children (truth-teller, liar, alternater-starts-truth, alternater-starts-lie) and whose columns are the three questions. Tool #2 (Systematic List) drives row-by-row Yes/No filling — for each row we just apply the rule mechanically. Tool #7 (Subproblems) breaks the work into (a) build the grid, (b) translate Yes-counts into equations, (c) solve. Tool #13 (Algebra) lets us subtract equations to isolate the truth-teller count without solving the whole system. The final candy count is just (number of truth-tellers) x (Yes answers each gives), and the row for truth-tellers in the grid will read Yes-No-No, so each contributes exactly one candy.

Execute — Answer: A

#4 Use Matrix Logic K.MD.B.3 Step 1

Build the grid (Tool #4).
Rows: T = truth-teller, L = liar, $A_t$ = alternater starting truth, $A_\ell$ = alternater starting lie.
Columns: Q1, Q2, Q3.
For each cell, ask "Is the child's spoken answer Yes or No?".
A truth-teller says the true statement; a liar says the negation.
Filling row by row gives the response table.

$$\begin{array}{c|ccc} & Q1 & Q2 & Q3 \\ \hline T & \text{Y} & \text{N} & \text{N} \\ L & \text{Y} & \text{Y} & \text{N} \\ A_t & \text{N} & \text{N} & \text{N} \\ A_\ell & \text{Y} & \text{Y} & \text{Y} \end{array}$$

💡 Kindergarten-level classification: sort children into categories and count Yes per row — the only "hard" part is reading the rules out loud one row at a time.

#2 Make a Systematic List K.MD.B.3 Step 2

Spot-check the trickiest row, $A_\ell$ (alternater starting with a lie).
Q1 "Are you a truth-teller?" — truth would be No, this child lies first, so says Yes.
Q2 "Are you an alternater?" — truth, this child now tells the truth, says Yes.
Q3 "Are you a liar?" — truth would be No, child lies again, says Yes.
All three Yes.

$$A_\ell: \;\text{Y}, \text{Y}, \text{Y}$$

💡 Just walk through Yes/No one question at a time using each child's rule — Kindergarten classification, no arithmetic yet.

#7 Identify Subproblems 6.EE.B.6 Step 3

Read the Yes columns.
Q1 "Yes" comes from T, L, $A_\ell$ -> $T + L + A_\ell = 22$.
Q2 "Yes" comes from L, $A_\ell$ -> $L + A_\ell = 15$.
Q3 "Yes" comes only from $A_\ell$ -> $A_\ell = 9$.

$$T + L + A_\ell = 22, \quad L + A_\ell = 15, \quad A_\ell = 9$$

💡 Each Yes count is just "add up the rows that said Yes in that column" — Grade 6 expression-writing turns the grid into three equations.

#13 Convert to Algebra 6.EE.B.7 Step 4

Tool #13 (algebra) on the cleanest move: subtract equation 2 from equation 1 to kill $L$ and $A_\ell$ at once.
The number of truth-tellers $T$ falls out directly.

$$(T + L + A_\ell) - (L + A_\ell) = 22 - 15 \;\Rightarrow\; T = 7$$

💡 One subtraction isolates $T$ — Grade 6 equation-solving avoids touching the other unknowns.

#4 Use Matrix Logic 3.OA.A.3 Step 5

Count the candy.
Look at the T-row of the grid: Yes-No-No, so every truth-teller earns exactly $1$ candy (only from Q1).
With $T = 7$ truth-tellers, total candy to truth-tellers $= 7 \times 1 = 7$.

$$\text{candy to truth-tellers} = T \times 1 = 7 \times 1 = 7$$

💡 Multiply candies-per-child by number of truth-tellers — Grade 3 multiplication.

#3 Eliminate Possibilities K.MD.B.3 Step 6

Match $7$ to the choices: it's option (A).

$$\text{answer} = 7 \;\Rightarrow\; \textbf{(A)}$$

💡 Final compare-to-options — the smallest, cleanest number on the list.

[1] #4 K.MD.B.3 Build the grid (Tool #4). Rows: T = truth-teller, L = liar, $A_t$ = alternater s

[2] #2 K.MD.B.3 Spot-check the trickiest row, $A_\ell$ (alternater starting with a lie). Q1 "Are

[3] #7 6.EE.B.6 Read the Yes columns. Q1 "Yes" comes from T, L, $A_\ell$ -> $T + L + A_\ell = 22

[4] #13 6.EE.B.7 Tool #13 (algebra) on the cleanest move: subtract equation 2 from equation 1 to

[5] #4 3.OA.A.3 Count the candy. Look at the T-row of the grid: Yes-No-No, so every truth-teller

[6] #3 K.MD.B.3 Match $7$ to the choices: it's option (A).

Review

Reasonableness: Sanity check the whole population. From the equations $A_\ell = 9$, $L = 15 - 9 = 6$, $T = 7$, so $A_t = 31 - 7 - 6 - 9 = 9$. That's $9$ alternaters of each starting type — perfectly reasonable. Re-totalling the Yes columns: Q1 Yes from $T + L + A_\ell = 7 + 6 + 9 = 22$ ✓. Q2 Yes from $L + A_\ell = 6 + 9 = 15$ ✓. Q3 Yes from $A_\ell = 9$ ✓. All three counts match the given data, so the model is consistent.

Alternative: Tool #16 (Change Focus / Complement): instead of finding $T$ directly, notice that the difference $Q1 - Q2 = 22 - 15 = 7$ exactly counts the rows that said Yes to Q1 but No to Q2. Looking at the grid, only the truth-tellers fit that signature, so $T = 7$ and the truth-teller candy is also $7$. Same answer (A), reached by complementary subtraction in one step.

CCSS standards used (min grade 6)

K.MD.B.3 Classify objects into given categories and count the numbers in each (Sorting the $31$ children into the four behavior categories (T, L, $A_t$, $A_\ell$) and filling a Yes/No grid row by row.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Multiplying the count of truth-tellers ($7$) by the candies each one earns ($1$) to get the total candy ($7$).)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $T, L, A_\ell$ stand for the populations of each behavior and translating each Yes column into a sum expression.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Subtracting $L + A_\ell = 15$ from $T + L + A_\ell = 22$ to isolate $T = 7$ in one move.)

⭐ This AMC 10 problem only needs Grade 6 expression-writing you already know — build a tiny Yes/No grid for the four kinds of children, read off the three Yes-counts as equations, and one subtraction ($22 - 15$) gives the truth-teller count directly. They say Yes only once, so the candy total is exactly $7$.