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Difference between revisions of "2000 AMC 12 Problems/Problem 8"

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[[Image:2000_12_AMC-8.png]]
 
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== Solution ==
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== Solution 1==
 
By counting the squares starting from the center of each figure, the figure 0 has 1 square, the figure 1 has <math>1 + 4(1)</math> squares, figure 2 has <math>1+4(1+2)</math> squares, and so on. Figure 100 would have <math>1 + 4(1 + 2 + \cdots + 100) = 1 + 4 \frac{100(101)}{2} = 20201 \Rightarrow \mathrm{(C)}</math>.
 
By counting the squares starting from the center of each figure, the figure 0 has 1 square, the figure 1 has <math>1 + 4(1)</math> squares, figure 2 has <math>1+4(1+2)</math> squares, and so on. Figure 100 would have <math>1 + 4(1 + 2 + \cdots + 100) = 1 + 4 \frac{100(101)}{2} = 20201 \Rightarrow \mathrm{(C)}</math>.
  
  
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Alternate solution:
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== 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 = <math>N</math>, note that <math>N^2 + (N+1)^2</math> represents the number of squares in the figure. For example, figure 4 has <math>4^2+5^2 = 41</math> squares. Therefore, the number of squares in figure 100 has <math>100^2 + 101^2 = 20201 \Rightarrow\mathrm{(C)}</math>.
 
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 = <math>N</math>, note that <math>N^2 + (N+1)^2</math> represents the number of squares in the figure. For example, figure 4 has <math>4^2+5^2 = 41</math> squares. Therefore, the number of squares in figure 100 has <math>100^2 + 101^2 = 20201 \Rightarrow\mathrm{(C)}</math>.
  

Revision as of 23:56, 31 May 2011

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 $100$?

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

2000 12 AMC-8.png

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 $\mathrm{C}$.

See also

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
All AMC 12 Problems and Solutions
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