Difference between revisions of "2000 AMC 12 Problems/Problem 8"
(solution) |
|||
Line 8: | Line 8: | ||
== Solution == | == Solution == | ||
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>. | ||
+ | |||
+ | |||
+ | Alternate solution: | ||
+ | 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>. | ||
== See also == | == See also == |
Revision as of 21:27, 11 September 2008
Problem
Figures , , , and consist of , , , and non-overlapping squares. If the pattern continued, how many non-overlapping squares would there be in figure ?
Solution
By counting the squares starting from the center of each figure, the figure 0 has 1 square, the figure 1 has squares, figure 2 has squares, and so on. Figure 100 would have .
Alternate solution:
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 = , note that represents the number of squares in the figure. For example, figure 4 has squares. Therefore, the number of squares in figure 100 has .
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 |