AMC 8 · 2023 · #8

Easy mode Grade 3

Problem

Imagine four friends — Lola, Lolo, Tiya, and Tiyo — playing in a ping pong tournament.

Every pair of friends plays each other exactly two times. So each player plays $6$ matches in total.

For each player, we write down a string of six $1$ s and $0$ s, one digit per match. A $1$ means that player won that match. A $0$ means they lost.

Here is the record so far. Tiyo's record is missing.

$\begin{tabular}{c | c} Player & Result \\ \hline Lola & \texttt{111011}\\ Lolo & \texttt{101010}\\ Tiya & \texttt{010100}\\ Tiyo & \texttt{??????} \end{tabular}$

(For example, Lola has five $1$ s, so Lola won five matches. The single $0$ is in the fourth spot, so Lola lost her fourth match.)

What was Tiyo's win-loss record?

Pick an answer.

(A)

$\texttt{000101}$

(B)

$\texttt{001001}$

(C)

$\texttt{010000}$

(D)

$\texttt{010101}$

(E)

$\texttt{011000}$

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Lola, Lolo, Tiya, and Tiyo each play every other player exactly twice, giving six rounds of matches. The table records each player's six results as a string of 1s (win) and 0s (loss). Lola = 111011, Lolo = 101010, Tiya = 010100, and Tiyo is unknown. We need to find Tiyo's six-digit win-loss string from the five answer choices.

Givens: Four players: Lola, Lolo, Tiya, Tiyo; Each pair of players meets exactly $2$ times; Lola's record: $\texttt{111011}$ (5 wins, 1 loss); Lolo's record: $\texttt{101010}$ (3 wins, 3 losses); Tiya's record: $\texttt{010100}$ (2 wins, 4 losses); Answer choices: (A) $\texttt{000101}$, (B) $\texttt{001001}$, (C) $\texttt{010000}$, (D) $\texttt{010101}$, (E) $\texttt{011000}$

Unknowns: Tiyo's six-digit win-loss string

Understand

Plan

Primary tool: #15 Organize Information in More Ways

Secondary: #3 Eliminate Possibilities, #2 Make a Systematic List

The table is naturally read row-by-row ("each player's record"), but the trick is to re-read it column-by-column — each column is one round of two simultaneous matches among the four players, so the four entries in any column must contain exactly two 1s (Tool #15). Once we know each column sums to 2, filling in Tiyo's row is just subtraction: column-sum minus the three known entries. We use Tool #3 (Eliminate Possibilities) as a sanity check against the listed choices, and Tool #2 (Make a Systematic List) to walk through the six columns in order without missing any.

Execute — Answer: A

#15 Organize Information in More Ways 3.OA.A.3 Step 1

Re-organize the table by COLUMNS instead of rows.
Four players play their matches at the same time, so each column of the table is one round.
With 4 players in a round, there are exactly $4 \div 2 = 2$ matches happening, which means exactly $2$ wins and $2$ losses per column.
So every column must contain exactly two $1$s and two $0$s.

$$\text{column sum} = 2 \text{ (for every column)}$$

💡 $4$ players grouped into pairs of opponents gives $2$ matches per round — basic Grade 3 multiplication/division reasoning.

#2 Make a Systematic List 1.OA.A.1 Step 2

Walk through the six columns systematically.
For each round, add Lola's, Lolo's, and Tiya's entries; whatever Tiyo needs to make the column sum equal to $2$ is Tiyo's result for that round.

$$\text{Tiyo's column} = 2 - (\text{Lola} + \text{Lolo} + \text{Tiya})$$

💡 Subtracting a known sum from $2$ to find the missing addend is a Grade 1 addition/subtraction-within-20 idea.

#2 Make a Systematic List 1.OA.A.1 Step 3

Column 1: Lola $1$ + Lolo $1$ + Tiya $0$ = $2$.
To keep the column sum at $2$, Tiyo $= 2 - 2 = 0$.
Column 2: $1 + 0 + 1 = 2$, so Tiyo $= 0$.
Column 3: $1 + 1 + 0 = 2$, so Tiyo $= 0$.
Column 4: $0 + 0 + 1 = 1$, so Tiyo $= 2 - 1 = 1$.
Column 5: $1 + 1 + 0 = 2$, so Tiyo $= 0$.
Column 6: $1 + 0 + 0 = 1$, so Tiyo $= 1$.

$$\text{Tiyo} = \texttt{0},\,\texttt{0},\,\texttt{0},\,\texttt{1},\,\texttt{0},\,\texttt{1}$$

💡 Each column needs the missing addend to make the sum $2$ — Grade 1 unknown-addend thinking, repeated six times.

#3 Eliminate Possibilities 1.OA.A.1 Step 4

Assemble Tiyo's six results left-to-right: $0,0,0,1,0,1 \;=\; \texttt{000101}$.
Compare to the answer choices: this matches (A) $\texttt{000101}$ exactly.
Eliminate the others: (C) $\texttt{010000}$ and (D) $\texttt{010101}$ would give the wrong total wins, and (B) $\texttt{001001}$ and (E) $\texttt{011000}$ disagree with at least one column constraint.

$$\texttt{000101} \;\Rightarrow\; \textbf{(A)}$$

💡 Matching a built-up answer string against the five choices is straightforward Grade 1 comparison.

[1] #15 3.OA.A.3 Re-organize the table by COLUMNS instead of rows. Four players play their matche

[2] #2 1.OA.A.1 Walk through the six columns systematically. For each round, add Lola's, Lolo's,

[3] #2 1.OA.A.1 Column 1: Lola $1$ + Lolo $1$ + Tiya $0$ = $2$. To keep the column sum at $2$, T

[4] #3 1.OA.A.1 Assemble Tiyo's six results left-to-right: $0,0,0,1,0,1 \;=\; \texttt{000101}$.

Review

Reasonableness: Cross-check by total wins. Across $4$ players and $6$ rounds with $2$ wins per round, the grand total of $1$s in the table must be $6 \times 2 = 12$. Lola has $5$, Lolo has $3$, Tiya has $2$, and our Tiyo $= \texttt{000101}$ has $2$, summing to $5 + 3 + 2 + 2 = 12$. The row total checks, the column constraint holds for every column, and choice (A) is the only one matching all six column equations.

Alternative: Alternative path with Tool #3 (Eliminate Possibilities) first: each player plays $6$ games, and the total wins must be $12$. Known wins are $5 + 3 + 2 = 10$, so Tiyo needs exactly $2$ wins — eliminating (C) (one $1$) and (D) (three $1$s) immediately. Then test the remaining three choices against the column-sum rule until only (A) survives. This is slightly less elegant than the column-by-column derivation but reaches the same answer.

CCSS standards used (min grade 3)

3.OA.A.3 Solve multiplication and division word problems within 100 (Realizing that $4$ players in a round split into $4 \div 2 = 2$ matches, so each round (column) contributes exactly $2$ wins.)
1.OA.A.1 Solve addition and subtraction word problems within 20 (Filling in Tiyo's missing entry per column as $2 - (\text{sum of the three known entries})$, and matching the final string $\texttt{000101}$ to the answer choices.)

⭐ This AMC 8 problem only needs Grade 3 ideas about splitting players into pairs of matches you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 3)

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