AMC 10 · 2013 · #6

Grade 1 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

logical-deductionparity casework ↑ Prerequisites: logical-deduction

📏 Medium solution 💡 2 insights

Problem

Joey and his five brothers are ages $3$ , $5$ , $7$ , $9$ , $11$ , and $13$ . One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home. How old is Joey?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2013 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.

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Toolkit + CCSS Solution

Understand

Restated: Six brothers (one of them Joey) have ages $3$, $5$, $7$, $9$, $11$, and $13$. Two of them whose ages add to $16$ went to the movies. Two more, each younger than $10$, went to play baseball. Joey and the $5$-year-old stayed home. Find Joey's age.

Givens: The six ages are $3, 5, 7, 9, 11, 13$; Two brothers went to the movies and their ages sum to $16$; Two brothers went to play baseball and each is younger than $10$; Joey and the $5$-year-old stayed home; Answer choices: (A) $3$, (B) $7$, (C) $9$, (D) $11$, (E) $13$

Unknowns: Joey's age

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #2 Make a Systematic List

The set of ages is small and fixed, and the answer is one of five choices — the classic trigger for Tool #3 (Eliminate Possibilities). First Tool #2 (Make a Systematic List) pins down every pair that could be the movie pair by adding to $16$; then Tool #3 knocks out the ones that clash with the other clues until only one arrangement survives. Joey is whoever is left standing.

Execute — Answer: D

#2 Make a Systematic List 1.OA.C.6 Step 1

List every pair that sums to 16

Go through the ages and find all pairs that add up to $16$, since that is the movie pair.
Checking the numbers, the only pairs are $3+13$, $5+11$, and $7+9$.

$$3+13=16,\quad 5+11=16,\quad 7+9=16$$

💡 Only a handful of pairs can add to $16$, so listing them all leaves nothing to guess about the movie pair.

#3 Eliminate Possibilities 1.OA.A.1 Step 2

Cross off the pair using the 5-year-old

The $5$-year-old stayed home, so the $5$-year-old cannot be at the movies.
That deletes the pair $5+11$.
The movie pair must be either $3+13$ or $7+9$.

$$\text{home} = \{5, \text{Joey}\} \Rightarrow 5 \notin \text{movies} \Rightarrow \text{movies} \in \{(3,13),\,(7,9)\}$$

💡 A brother can be in only one place, so a fact about home instantly removes a movie option.

#3 Eliminate Possibilities 1.NBT.B.3 Step 3

Test which movie pair leaves a legal baseball pair

Baseball needs two brothers younger than $10$, and the $5$-year-old is home, so the players come from $\{3, 7, 9\}$.
Suppose the movie pair were $7+9$: then baseball could only draw from $\{3\}$ — just one player, which is not enough.
So $7+9$ fails, and the movie pair must be $3+13$.

$$\text{under }10 = \{3,7,9\};\ \text{movies}=(7,9)\Rightarrow \text{baseball}\subseteq\{3\}\ (\text{too few}) \Rightarrow \text{movies}=(3,13)$$

💡 Baseball demands two young players, so any choice that starves it of a second young brother is impossible.

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

Place the rest; Joey is who remains

With the movies taking $3$ and $13$, the remaining under-$10$ brothers are $7$ and $9$, so they are the baseball pair.
Home already has the $5$-year-old plus Joey.
The ages $3, 5, 7, 9, 13$ are all assigned, leaving only $11$ for Joey.
So Joey is $11$, which is choice (D).

$$\text{movies}=\{3,13\},\ \text{baseball}=\{7,9\},\ \text{home}=\{5,\,\text{Joey}\}\Rightarrow \text{Joey}=11\ \textbf{(D)}$$

💡 Once five ages have homes, the single leftover age has to be Joey's.

[1] #2 1.OA.C.6 Go through the ages and find all pairs that add up to $16$, since that is the mo

[2] #3 1.OA.A.1 The $5$-year-old stayed home, so the $5$-year-old cannot be at the movies. That

[3] #3 1.NBT.B.3 Baseball needs two brothers younger than $10$, and the $5$-year-old is home, so

[4] #3 1.OA.A.1 With the movies taking $3$ and $13$, the remaining under-$10$ brothers are $7$ a

Review

Reasonableness: Check every clue against the final arrangement: movies $= 3+13 = 16$ (sum is $16$, good); baseball $= 7$ and $9$, both under $10$ (good); home $=$ the $5$-year-old and Joey ($11$). All six brothers $3,5,7,9,11,13$ are used exactly once, and each clue is satisfied, so Joey being $11$ is consistent.

Alternative: Tool #16 (Change Focus / Count the Complement): instead of tracking three groups, ask which single age is never forced anywhere. The movie clue can use $3,13,7,9$; the baseball clue can use $3,7,9$; the home clue names $5$. The one age that no clue can pull out of the house is $11$ — and since Joey is the un-named stay-home brother, Joey is $11$.

CCSS standards used (min grade 1)

1.OA.C.6 Add and subtract within 20 (Finding every pair of ages that sums to $16$ to pin down the possible movie pairs.)
1.OA.A.1 Use addition and subtraction within 20 to solve word problems (Translating the story clues (who is home, who is left over) into which brother goes where, ending with Joey.)
1.NBT.B.3 Compare two two-digit numbers using >, =, and < (Deciding which brothers are younger than $10$ so they qualify for the baseball pair.)

⭐ Write down every pair that fits a clue, cross off the ones the other clues forbid, and the brother with no group left is your answer.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

List every pair that sums to 16

Cross off the pair using the 5-year-old

Test which movie pair leaves a legal baseball pair

Place the rest; Joey is who remains

Review

CCSS standards used (min grade 1)

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