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2004 AMC 10A Problem 16

2004 AMC 10A Problem 16AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Counting · difficulty 3 of 5 · about 3 min · position P16-20

The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$ . How many of these squares contain the black center square?

A) $12$
B) $15$
C) $17$
D) $19$
E) $20$

Answer

Key insight

Count by size. The numbers of squares that cover the center are 1, 4, 9, 4, 1 for sizes 1 through 5, which adds to 19.

Solution

Count the squares one size at a time. A square is fixed once you know which rows and which columns it covers, and the black square sits in row $3$ , column $3$ .

1. Rows first. For each size, list the groups of neighboring rows that include row $3$ .
- Size $1$ : just row $3$ . That is $1$ choice.
- Size $2$ : rows $2$ – $3$ or $3$ – $4$ . That is $2$ choices.
- Size $3$ : rows $1$ – $3$ , $2$ – $4$ or $3$ – $5$ . That is $3$ choices.
- Size $4$ : rows $1$ – $4$ or $2$ – $5$ . That is $2$ choices.
- Size $5$ : rows $1$ – $5$ . That is $1$ choice.
2. Columns. The columns work exactly the same way, so each size has the same number of column choices.
3. Squares. Any row choice can go with any column choice, so for each size multiply the row count by itself:

size	row choices	squares with the center
$1 \times 1$	1	$1$
$2 \times 2$	2	$4$
$3 \times 3$	3	$9$
$4 \times 4$	2	$4$
$5 \times 5$	1	$1$

Total: $1 + 4 + 9 + 4 + 1 = 19$ .

The answer is $\boxed{\textbf{(D)}\ 19}$ .

Why this works

A square in a grid is decided by its rows and its columns, and those two choices are separate, so each count is a number times itself. Notice the pattern $1, 4, 9, 4, 1$ : small squares are limited because they must reach the center, and big ones because they must fit in the grid. That balance is a good self-check.

Alternative approach

Count the squares that miss the center and subtract.

The grid has $25 + 16 + 9 + 4 + 1 = 55$ squares of all sizes.
Every square of size $3$ or more covers the center, so only small squares can miss it.
Of the $25$ unit squares, $24$ miss it. Of the $16$ squares of size $2$ , the $4$ found above cover it, so $12$ miss it.
That leaves $55 - 24 - 12 = 19$ .

The trap

Counting every square in the grid (55), or forgetting the 4 by 4 squares and getting 15.

Concise solution

Number the rows and columns $1$ through $5$ ; the black cell is at row $3$ , column $3$ . A $k \times k$ square is determined by its top row $i$ and left column $j$ , each ranging over $1, \dots, 6 - k$ . It contains the center exactly when $i \le 3 \le i + k - 1$ and the same for $j$ ; by symmetry the row and column counts are equal, so the answer for size $k$ is (number of valid $i$ ) $^2$ .

$k = 1$ : $i = 3$ only. $1^2 = 1$ .
$k = 2$ : $i \in \{2, 3\}$ . $2^2 = 4$ .
$k = 3$ : $i \in \{1, 2, 3\}$ . $3^2 = 9$ .
$k = 4$ : $i \in \{1, 2\}$ (top row at most $2$ ). $2^2 = 4$ .
$k = 5$ : $i = 1$ only. $1^2 = 1$ .

Total: $1 + 4 + 9 + 4 + 1 = 19$ .

The answer is $\boxed{\textbf{(D)}\ 19}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 5: More With Combinations (§5.2)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)

Techniques

Split into exhaustive cases and handle each · Exploit symmetry to reduce work or pair up objects