AMC 8 · 2004 · #18
Grade 3 countingProblem
Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers 1 through 10. Each throw hits the target in a region with a different value. The scores are: Alice 16 points, Ben 4 points, Cindy 7 points, Dave 11 points, and Ellen 17 points. Who hits 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.
More like this
Same archetype — closest grade level first.
