2013 AMC 10A Problem 6

Start with the clue that gives the most information: the pairs that add to $16$.

1. Pairs of ages that add to 16. From $3, 5, 7, 9, 11, 13$:
$$3 + 13 = 16, \qquad 5 + 11 = 16, \qquad 7 + 9 = 16.$$
2. Rule one out. The $5$-year-old stayed home, so he did not go to the movies. The pair $5, 11$ is out.
3. Try 7 and 9 at the movies. The boys under $10$ are $3, 5, 7, 9$. With $7$ and $9$ at the movies and $5$ at home, only the $3$-year-old is left. Baseball needs two boys under $10$, so this fails.
4. Try 3 and 13 at the movies. Now $7$ and $9$ are free, and both are under $10$, so they play baseball. The $5$-year-old is home.
5. Who is left? The ages used so far are $3, 13, 7, 9, 5$. The only age left is $11$, and that must be Joey, who stayed home.

Joey is $\boxed{\textbf{(D)}\ 11}$.

In "who went where" puzzles, list the few choices the strongest clue allows, then test each one against every other clue. One clue that fails throws out the whole choice. Here only two cases were left after the first clue, so checking both was quick.

Pairs of ages from $\{3, 5, 7, 9, 11, 13\}$ that sum to $16$: $(3, 13)$, $(5, 11)$, $(7, 9)$.

The $5$-year-old stayed home, so the movie pair is not $(5, 11)$. That leaves two cases.

Case $(7, 9)$ at the movies: the brothers under $10$ who remain are just the $3$-year-old (the $5$-year-old is home), so two of them cannot go to baseball. Impossible.

Case $(3, 13)$ at the movies: the remaining brothers under $10$ are $7$ and $9$, and they go to baseball. The one person left is $11$, who must be Joey.

Joey is $\boxed{\textbf{(D)}\ 11}$.