2004 AMC 10B Problem 1

P1-10~1 minPrint

Basic Counting Set up the equation/formula and compute; no special trick needed

Each row of the Misty Moon Amphitheater has $33$ seats. Rows $12$ through $22$ are reserved for a youth club. How many seats are reserved for this club?

Show solution without answering

Key insight

Rows 12 through 22 means both end rows count, so there are 11 rows, not 10. Then multiply 11 rows by 33 seats.

Solution

1. First count the rows. It is tempting to do $22 - 12 = 10$ , but try a small case to check that idea: rows $1$ through $3$ are rows $1, 2, 3$ , which is $3$ rows, while $3 - 1 = 2$ . Subtracting misses one row, because both end rows are included.
2. So rows $12$ through $22$ are $22 - 12 + 1 = 11$ rows. You can also see this by listing them: $12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22$ . That is $11$ labels.
3. Each row has $33$ seats, so the club gets
$11 \times 33 = 363$
seats.

The answer is $\boxed{\textbf{(C)}\ 363}$ .

Why this works

When a problem says "from $12$ through $22$ ," both ends are in the list, so the count is (last minus first) plus one. If you are ever unsure, test the rule on a tiny list you can count by hand, like $1$ through $3$ . Choice (B) $330$ is exactly the answer you get by forgetting the plus one.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

First count the rows. The labels $12, 13, \dots, 22$ form a list of $22 - 12 + 1 = 11$ numbers; one way to see this is to subtract $11$ from every label, which turns the list into $1, 2, \dots, 11$ .

Each of these $11$ rows holds $33$ seats, so the club gets
$11 \cdot 33 = 363$
seats.

The answer is $\boxed{\textbf{(C)}\ 363}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Counting & Probability, Chapter 4: Committees and Combinations