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2001 AMC 10 Problem 11

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2001 AMC 10 Problem 11AMC 10 Step by Step
Name:                             Date: September 13, 2026
Sequences & Series · difficulty 2 of 5 · about 2 min · position P11-15

Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains 88 unit squares. The second ring contains 1616 unit squares. If we continue this process, the number of unit squares in the 100th100^\text{th} ring is

figure

  1. A)

    396396

  2. B)

    404404

  3. C)

    800800

  4. D)

    10,00010, 000

  5. E)

    10,40410, 404

Answer

C

Key insight

The nth ring is a (2n+1)-square minus a (2n-1)-square, and (2n+1)^2 - (2n-1)^2 = 8n, so the 100th ring has 800 squares.

Solution

The center square together with the first nn rings forms a solid square of side 2n+12n + 1 (one ring adds one unit on each side). The nnth ring alone is that square with the side-(2n−1)(2n-1) square removed:
(2n+1)2−(2n−1)2=[(2n+1)−(2n−1)][(2n+1)+(2n−1)]=2⋅4n=8n.(2n+1)^2 - (2n-1)^2 = \big[(2n+1) - (2n-1)\big]\big[(2n+1) + (2n-1)\big] = 2 \cdot 4n = 8n.
Check against the given data: n=1n = 1 gives 88 and n=2n = 2 gives 1616.

For n=100n = 100 the ring contains 8⋅100=8008 \cdot 100 = 800 unit squares.

The answer is (C) 800\boxed{\textbf{(C)}\ 800}.

Why this works

Concentric layers around a center are differences of consecutive "solid" shapes, and a difference of squares factors into something linear. The ring count grows linearly (8,16,24,…8, 16, 24, \dots) even though the areas grow quadratically; recognizing that the two given values fit 8n8n, and confirming with the formula, avoids guessing.

Alternative approach

Count the ring directly: it has two horizontal strips of 2n+12n + 1 squares (top and bottom) and two vertical strips of 2n−12n - 1 squares (left and right, corners already used), totaling 2(2n+1)+2(2n−1)=8n2(2n+1) + 2(2n-1) = 8n.

The trap

Computing the whole 201 by 201 block (40401) or the block minus the center, instead of the single outer ring.

Common mistakes

  • Computing the whole 201 by 201 block (40401) or the block minus the center, instead of the single outer ring.
  • Using side length 2n2n instead of 2n+12n + 1, or double counting the four corner squares when adding the strips.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Compute small cases, spot the pattern, generalize