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2000 AMC 10 Problem 12

2000 AMC 10 Problem 12AMC 10 Step by Step

Name: Date: October 28, 2026

Sequences & Series · difficulty 2 of 5 · about 3 min · position P11-15

Figures $0$ , $1$ , $2$ , and $3$ consist of $1$ , $5$ , $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?

A) $10401$
B) $19801$
C) $20201$
D) $39801$
E) $40801$

Answer

Key insight

Figure n has n squared plus (n+1) squared unit squares, so figure 100 has 100 squared plus 101 squared, which is 20201.

Solution

1. Look at the counts: $1, 5, 13, 25$ . They grow by $4, 8, 12$ , and those jumps grow by $4$ each time. So the next jump is $16$ and figure $4$ has $41$ squares.
2. Try to spot a pattern using square numbers:
$1 = 0^2 + 1^2,\quad 5 = 1^2 + 2^2,\quad 13 = 2^2 + 3^2,\quad 25 = 3^2 + 4^2.$
Figure $n$ seems to have $n^2 + (n+1)^2$ squares.
3. Check it on figure $4$ , which we found has $41$ squares: $4^2 + 5^2 = 16 + 25 = 41$ . It matches.
4. The pattern also matches the picture. Read figure $n$ row by row: the rows have $1, 3, 5, \ldots, 2n+1, \ldots, 5, 3, 1$ squares. The top part down to the longest row is the first $n+1$ odd numbers, which add to $(n+1)^2$ . The part below is the first $n$ odd numbers, which add to $n^2$ . For figure $2$ : $1 + 3 + 5 = 9$ and $3 + 1 = 4$ , total $13$ .
5. For figure $100$ , use $n = 100$ (not $99$ , since the first figure is figure $0$ ):
$100^2 + 101^2 = 10000 + 10201 = 20201.$

The answer is $\boxed{\textbf{(C)}\ 20201}$ .

Why this works

When the jumps between terms themselves grow by the same amount each time, the pattern is a "squared" formula, and small cases usually show it quickly. Always check a guessed formula on one more term before using it for $n = 100$ , and watch whether the list starts at $0$ or $1$ .

Alternative approach

Count the picture directly. Each figure has one center square and four arms. In figure $n$ , each arm is a staircase with $1 + 2 + \cdots + n$ squares. For $n = 100$ , one arm has $\frac{100 \cdot 101}{2} = 5050$ squares, so the total is $1 + 4 \cdot 5050 = 20201$ .

The trap

Counting wrong because the figures start at 0: using 99 squared plus 100 squared = 19801 gives the wrong choice (B).

Concise solution

The counts $1, 5, 13, 25$ have first differences $4, 8, 12$ , which themselves go up by a constant $4$ . Constant second differences mean the count is a quadratic in the figure number $n$ .

Rather than solve for coefficients, match the data to a recognizable form:
$1 = 0^2 + 1^2,\quad 5 = 1^2 + 2^2,\quad 13 = 2^2 + 3^2,\quad 25 = 3^2 + 4^2.$
So figure $n$ contains $n^2 + (n+1)^2$ squares. As a check, the next difference should be $16$ , and $4^2 + 5^2 = 41 = 25 + 16$ .

For $n = 100$ :
$100^2 + 101^2 = 10000 + 10201 = 20201.$

The answer is $\boxed{\textbf{(C)}\ 20201}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)

Techniques

Compute small cases, spot the pattern, generalize