Difference between revisions of "2000 AMC 12 Problems/Problem 8"
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Revision as of 04:48, 3 November 2021
- The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.
Contents
[hide]Problem
Figures , , , and consist of , , , and nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
Video:
https://www.youtube.com/watch?v=HVP6qjKAkjA&t=2s
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Solution 3
We can see that each figure has a central box and 4 columns of boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are squares. . Adding in the original center box we have .
Solution 4
Let be the number of squares in figure . We can easily see that Note that in , the number multiplied by the 4 is the th triangular number. Hence, . ~qkddud~
Solution 5
Let denote the number of unit cubes in a figure. We have
Computing the difference between the number of cubes in each figure yields It is easy to notice that this is an arithmetic sequence, with the first term being and the difference being . Let this sequence be
From to , the sequence will have terms. Using the arithmetic sum formula yields
So unit cubes.
~ljlbox
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 |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.