AMC 8 · 2004 · #18
Easy mode Grade 3Problem
Imagine a dartboard with 10 regions on it. The regions are worth 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 points.
Five friends take turns throwing darts: Alice, Ben, Cindy, Dave, and Ellen. Each friend throws 2 darts. Each person's total score is the sum of the two regions they hit.
Every dart hits a different region. So all 10 regions get hit exactly once.
Here are the totals:
- Alice scored 16 points.
- Ben scored 4 points.
- Cindy scored 7 points.
- Dave scored 11 points.
- Ellen scored 17 points.
Which friend hit the region worth 6 points?
Pick an answer.
AMC 8 2004 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.
Toolkit + CCSS Solution
Understand
Restated: Five friends — Alice, Ben, Cindy, Dave, Ellen — each throw two darts at a target whose regions are worth $1$ through $10$ points. Across all $10$ throws, every region is hit exactly once. Each person's score is the sum of their two regions. Alice scored $16$, Ben $4$, Cindy $7$, Dave $11$, and Ellen $17$. Who hit the region worth $6$?
Givens: Regions on the target are the whole numbers $1, 2, 3, \ldots, 10$; Five friends throw two darts each, so $10$ darts in total; Every region is hit exactly once (each value from $1$ to $10$ is used by exactly one dart); Scores: Alice $=16$, Ben $=4$, Cindy $=7$, Dave $=11$, Ellen $=17$; Answer choices: (A) Alice, (B) Ben, (C) Cindy, (D) Dave, (E) Ellen
Unknowns: Which friend's two-dart pair contains the region worth $6$
Understand
Restated: Five friends — Alice, Ben, Cindy, Dave, Ellen — each throw two darts at a target whose regions are worth $1$ through $10$ points. Across all $10$ throws, every region is hit exactly once. Each person's score is the sum of their two regions. Alice scored $16$, Ben $4$, Cindy $7$, Dave $11$, and Ellen $17$. Who hit the region worth $6$?
Givens: Regions on the target are the whole numbers $1, 2, 3, \ldots, 10$; Five friends throw two darts each, so $10$ darts in total; Every region is hit exactly once (each value from $1$ to $10$ is used by exactly one dart); Scores: Alice $=16$, Ben $=4$, Cindy $=7$, Dave $=11$, Ellen $=17$; Answer choices: (A) Alice, (B) Ben, (C) Cindy, (D) Dave, (E) Ellen
Plan
Primary tool: #4 Use Matrix Logic
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
This is a classic "who has what?" puzzle: five people, ten numbered regions, and the constraint that each region is used exactly once. Tool #4 (Use Matrix Logic) is the natural primary — we track which numbers each person could still own and force assignments as evidence builds. To feed the matrix, Tool #2 (Make a Systematic List) writes out every distinct pair $\{a,b\}$ with $a+b$ equal to each score. Tool #3 (Eliminate Possibilities) then crosses off any pair containing a number that some earlier assignment has already locked in. Starting from the person with the fewest possible pairs (Ben, who has only one) cascades to a unique solution.
Execute — Answer: A
2.OA.B.2 Step 1 - List every distinct pair of regions $1$–$10$ whose sum matches each friend's score.
- This is the candidate matrix.
💡 Grade 2 "add and subtract within $20$" is exactly the arithmetic needed — and listing all pairs systematically (smallest first) makes sure no case is missed.
3.OA.D.8 Step 2 - Lock in Ben.
- His score $4$ has only one valid pair $\{1,3\}$, so Ben must have hit regions $1$ and $3$.
- Mark $1$ and $3$ as used and cross out any pair in the matrix that contains them.
💡 Grade 3 multi-step reasoning: a single forced assignment ripples outward through the constraint that each region is used only once.
3.OA.D.8 Step 3 - Lock in Cindy.
- After Ben's elimination, only $\{2,5\}$ remains for Cindy.
- So Cindy hit regions $2$ and $5$.
- Update the used set and eliminate any pair containing $2$ or $5$.
💡 Elimination shrinks Cindy's row to a single survivor, the matrix-logic move.
3.OA.D.8 Step 4 - Lock in Dave.
- Of Dave's original five pairs, only $\{4,7\}$ avoids $\{1,2,3,5\}$.
- So Dave hit $4$ and $7$.
💡 Crossing out four of Dave's five pairs leaves exactly one — another forced row.
3.OA.D.8 Step 5 - Finish Alice and Ellen from the leftover set $\{6,8,9,10\}$.
- Check the three possible sums: $6+8=14$, $6+9=15$, $6+10=16$, $8+9=17$, $8+10=18$, $9+10=19$.
- Only $6+10=16$ matches Alice's score, and only $8+9=17$ matches Ellen's.
💡 With only four numbers left and two target sums, the pairing is forced — the matrix is complete.
3.OA.D.8 Step 6 - Read off who hit the $6$.
- The region $6$ appears in Alice's pair $\{6,10\}$.
💡 Once the matrix is filled, the question reduces to a one-line lookup.
2.OA.B.2 List every distinct pair of regions $1$–$10$ whose sum matches each friend's sco 3.OA.D.8 Lock in Ben. His score $4$ has only one valid pair $\{1,3\}$, so Ben must have h 3.OA.D.8 Lock in Cindy. After Ben's elimination, only $\{2,5\}$ remains for Cindy. So Cin 3.OA.D.8 Lock in Dave. Of Dave's original five pairs, only $\{4,7\}$ avoids $\{1,2,3,5\}$ 3.OA.D.8 Finish Alice and Ellen from the leftover set $\{6,8,9,10\}$. Check the three pos 3.OA.D.8 Read off who hit the $6$. The region $6$ appears in Alice's pair $\{6,10\}$. Review
Reasonableness: Verify the full assignment satisfies every condition. Ben $1+3=4$ ✓, Cindy $2+5=7$ ✓, Dave $4+7=11$ ✓, Alice $6+10=16$ ✓, Ellen $8+9=17$ ✓. The regions used are $\{1,3,2,5,4,7,6,10,8,9\}=\{1,2,\ldots,10\}$ — every region hit exactly once ✓. Also, $1+2+\cdots+10=55$ and $4+7+11+16+17=55$ ✓. The other answer choices fail: Ben's pair is $\{1,3\}$ (no $6$), Cindy's is $\{2,5\}$ (no $6$), Dave's is $\{4,7\}$ (no $6$), Ellen's is $\{8,9\}$ (no $6$). Only Alice owns the region worth $6$, so (A) is forced.
Alternative: Tool #16 (Change Focus / Count the Complement): instead of building up, work down by asking "who is the $6$ paired with?" If $6$ is in Cindy's pair, the partner is $1$ (since $6+1=7$); but then Ben (only pair $\{1,3\}$) needs $1$, contradiction. If $6$ is in Dave's pair, the partner is $5$ (since $5+6=11$); but Cindy then loses $\{2,5\}$ as well as $\{1,6\}$ and $\{3,4\}$ (those use $1$ or $3$ from Ben), leaving Cindy no valid pair — contradiction. So $6$ must belong to Alice (partner $10$, since $6+10=16$). Same answer (A) by direct contradiction.
CCSS standards used (min grade 3)
2.OA.B.2Fluently add and subtract within $20$ using mental strategies (Listing every pair of distinct whole numbers from $1$ to $10$ that adds to each given score ($4$, $7$, $11$, $16$, $17$).)3.OA.D.8Solve two-step word problems using the four operations; assess the reasonableness of answers (Chaining the deductions "Ben locks $\{1,3\}$ → Cindy locks $\{2,5\}$ → Dave locks $\{4,7\}$ → Alice/Ellen split $\{6,8,9,10\}$" to reach the final assignment, and checking the answer by re-summing each pair.)
⭐ When every clue is a sum and every number is used exactly once, start with the person whose score has only one possible pair — that one fact unlocks the next, and the rest falls like dominoes.
⭐ When every clue is a sum and every number is used exactly once, start with the person whose score has only one possible pair — that one fact unlocks the next, and the rest falls like dominoes.
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