2000 AMC 12 Problems/Problem 8

Problem

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

$\text {(A)}\ 10401 \qquad \text {(B)}\ 19801 \qquad \text {(C)}\ 20201 \qquad \text {(D)}\ 39801 \qquad \text {(E)}\ 40801$

Solution 1

By counting the squares starting from the center of each figure, the figure 0 has 1 square, the figure 1 has $1 + 4(1)$ squares, figure 2 has $1+4(1+2)$ squares, and so on. Figure 100 would have $1 + 4(1 + 2 + \cdots + 100) = 1 + 4 \frac{100(101)}{2} = 20201 \Rightarrow \mathrm{(C)}$ .

Solution 2

Note that figure 0 has 1 square, figure 1 has 5 squares, figure 2 has 13 squares, and so on. If we let the number of the figure = $N$ , note that $N^2 + (N+1)^2$ represents the number of squares in the figure. For example, figure 4 has $4^2+5^2 = 41$ squares. Therefore, the number of squares in figure 100 has $100^2 + 101^2 = 20201 \Rightarrow\mathrm{(C)}$ .

$2^{\text{nd}}$ alternate solution: For the $n^{\text{th}}$ figure, note that it could be constructed by making a $(2n+1)\times (2n+1)$ square, and then removing the $n^{\text{th}}$ triangular number from each of its corners. So, if $a_n$ represents the amount of squares in figure $n$ , $a_{n} = (2n+1)^2-4\frac{(n)(n+1)}{2}= 2n^2+2n+1$ . Therefore, $a_{100} = 20201$ , which gives $\mathrm{C}$ .

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2000 AMC 12 Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See also