← Sensim Lab Sensim Math

AMC 8 · 2004 · #18

Easy mode Grade 3
📗 View original problem →

Problem

Imagine a dartboard with 1010 regions on it. The regions are worth 11, 22, 33, 44, 55, 66, 77, 88, 99, and 1010 points.

Five friends take turns throwing darts: Alice, Ben, Cindy, Dave, and Ellen. Each friend throws 22 darts. Each person's total score is the sum of the two regions they hit.

Every dart hits a different region. So all 1010 regions get hit exactly once.

Here are the totals:

  • Alice scored 1616 points.
  • Ben scored 44 points.
  • Cindy scored 77 points.
  • Dave scored 1111 points.
  • Ellen scored 1717 points.

Which friend hit the region worth 66 points?

Pick an answer.

(A)
Alice
(B)
Ben
(C)
Cindy
(D)
Dave
(E)
Ellen
View mode:

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 Make a Systematic List 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.
$$\begin{array}{l|l}\text{Score} & \text{Possible pairs}\\\hline \text{Ben }4 & \{1,3\}\\ \text{Cindy }7 & \{1,6\},\,\{2,5\},\,\{3,4\}\\ \text{Dave }11 & \{1,10\},\,\{2,9\},\,\{3,8\},\,\{4,7\},\,\{5,6\}\\ \text{Alice }16 & \{6,10\},\,\{7,9\}\\ \text{Ellen }17 & \{7,10\},\,\{8,9\}\end{array}$$

💡 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.

#4 Use Matrix Logic 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.
$\text{Ben}=\{1,3\}\;\Rightarrow\;\text{used}=\{1,3\}$. Cross out $\{1,6\}$ and $\{3,4\}$ from Cindy; cross out $\{1,10\}$ and $\{3,8\}$ from Dave.

💡 Grade 3 multi-step reasoning: a single forced assignment ripples outward through the constraint that each region is used only once.

#3 Eliminate Possibilities 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 $5$.
$\text{Cindy}=\{2,5\}\;\Rightarrow\;\text{used}=\{1,2,3,5\}$. Cross out $\{2,9\}$ and $\{5,6\}$ from Dave.

💡 Elimination shrinks Cindy's row to a single survivor, the matrix-logic move.

#3 Eliminate Possibilities 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 $7$.
$\text{Dave}=\{4,7\}\;\Rightarrow\;\text{used}=\{1,2,3,4,5,7\}$. Remaining regions: $\{6,8,9,10\}$.

💡 Crossing out four of Dave's five pairs leaves exactly one — another forced row.

#4 Use Matrix Logic 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.
$$\text{Alice}=\{6,10\},\;\;\text{Ellen}=\{8,9\}$$

💡 With only four numbers left and two target sums, the pairing is forced — the matrix is complete.

#4 Use Matrix Logic 3.OA.D.8 Step 6
  • Read off who hit the $6$.
  • The region $6$ appears in Alice's pair $\{6,10\}$.
$$6 \in \text{Alice's pair} \;\Rightarrow\; \textbf{(A)}\ \text{Alice}$$

💡 Once the matrix is filled, the question reduces to a one-line lookup.

[1] #2 2.OA.B.2 List every distinct pair of regions $1$–$10$ whose sum matches each friend's sco
[2] #4 3.OA.D.8 Lock in Ben. His score $4$ has only one valid pair $\{1,3\}$, so Ben must have h
[3] #3 3.OA.D.8 Lock in Cindy. After Ben's elimination, only $\{2,5\}$ remains for Cindy. So Cin
[4] #3 3.OA.D.8 Lock in Dave. Of Dave's original five pairs, only $\{4,7\}$ avoids $\{1,2,3,5\}$
[5] #4 3.OA.D.8 Finish Alice and Ellen from the leftover set $\{6,8,9,10\}$. Check the three pos
[6] #4 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.2 Fluently 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.8 Solve 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.