AMC 10 Step by Step
Log inSign up
Printable

2004 AMC 10B Problem 1

Problem onlyBack to the problem
2004 AMC 10B Problem 1AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Counting · difficulty 1 of 5 · about 1 min · position P1-10

Each row of the Misty Moon Amphitheater has 3333 seats. Rows 1212 through 2222 are reserved for a youth club. How many seats are reserved for this club?

  1. A)

    297297

  2. B)

    330330

  3. C)

    363363

  4. D)

    396396

  5. E)

    726726

Answer

C

Key insight

Rows 12 through 22 inclusive are 22 - 12 + 1 = 11 rows, not 10, so the reserved block is 11 times 33.

Solution

First count the rows. The labels 12,13,…,2212, 13, \dots, 22 form a list of 22−12+1=1122 - 12 + 1 = 11 numbers; one way to see this is to subtract 1111 from every label, which turns the list into 1,2,…,111, 2, \dots, 11.

Each of these 1111 rows holds 3333 seats, so the club gets
11⋅33=36311 \cdot 33 = 363
seats.

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

Why this works

A range of consecutive integers from mm to nn contains n−m+1n - m + 1 members, because both ends are counted. This "fencepost" adjustment shows up whenever a problem says "from ... through ..."; the safe habit is to shift the range so it starts at 11 and read off the count.

The trap

Subtracting 22 - 12 = 10 rows and answering 330, forgetting that both endpoint rows are included.

Common mistakes

  • Subtracting 22 - 12 = 10 rows and answering 330, forgetting that both endpoint rows are included.
  • Multiplying 33⋅22=72633 \cdot 22 = 726 (choice (E)), which counts every row up to row 2222 rather than only the reserved ones.

Techniques

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