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2013 AMC 10A Problem 6

2013 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Logic Puzzles · difficulty 2 of 5 · about 2 min · position P1-10

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?

A)
$3$
B)
$7$
C)
$9$
D)
$11$
E)
$13$

Answer

Key insight

With the 5-year-old home, the movie pair is 3 and 13 or 7 and 9; only 3 and 13 leaves two under-10 brothers for baseball, so Joey is 11.

Solution

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

Why this works

Assignment puzzles are solved by starting with the most restrictive clue (here, a specific person is already placed at home) and then testing the few remaining possibilities against every other clue. Each candidate must satisfy all conditions simultaneously; one contradiction eliminates the whole case.

The trap

Choosing the movie pair 7 and 9 without checking that two brothers under 10 must still be free for baseball.

Common mistakes

Choosing the movie pair 7 and 9 without checking that two brothers under 10 must still be free for baseball.
Forgetting that Joey himself is one of the six ages, and so must be whoever is left after the four other brothers are placed.

Techniques

Split into exhaustive cases and handle each