Difference between revisions of "2000 AMC 12 Problems/Problem 8"
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<math>\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801</math> | <math>\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801</math> | ||
− | ===Solution 1 | + | ==Video:== |
+ | https://www.youtube.com/watch?v=HVP6qjKAkjA&t=2s | ||
+ | |||
+ | ==Solution 1== | ||
We can divide up figure <math>n</math> to get the sum of the sum of the first <math>n+1</math> odd numbers and the sum of the first <math>n</math> odd numbers. If you do not see this, here is the example for <math>n=3</math>: | We can divide up figure <math>n</math> to get the sum of the sum of the first <math>n+1</math> odd numbers and the sum of the first <math>n</math> odd numbers. If you do not see this, here is the example for <math>n=3</math>: | ||
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The sum of the first <math>n</math> odd numbers is <math>n^2</math>, so for figure <math>n</math>, there are <math>(n+1)^2+n^2</math> unit squares. We plug in <math>n=100</math> to get <math>\boxed{\textbf{(C) }20201}</math>. | The sum of the first <math>n</math> odd numbers is <math>n^2</math>, so for figure <math>n</math>, there are <math>(n+1)^2+n^2</math> unit squares. We plug in <math>n=100</math> to get <math>\boxed{\textbf{(C) }20201}</math>. | ||
− | + | ==Solution 2== | |
Using the recursion from solution 1, we see that the first differences of <math>4, 8, 12, ...</math> form an arithmetic progression, and consequently that the second differences are constant and all equal to <math>4</math>. Thus, the original sequence can be generated from a quadratic function. | Using the recursion from solution 1, we see that the first differences of <math>4, 8, 12, ...</math> form an arithmetic progression, and consequently that the second differences are constant and all equal to <math>4</math>. Thus, the original sequence can be generated from a quadratic function. | ||
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Calculating the answer to our problem, <math>f(100) = 20000 + 200 + 1 = 20201</math>, which is choice <math>\boxed{\textbf{(C) }20201}</math>. | Calculating the answer to our problem, <math>f(100) = 20000 + 200 + 1 = 20201</math>, which is choice <math>\boxed{\textbf{(C) }20201}</math>. | ||
− | + | ==Solution 3== | |
We can see that each figure <math>n</math> has a central box and 4 columns of <math>n</math> 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 <math>\sum_{n=1}^{100} n = 5050</math> squares. <math>4 \cdot 5050 = 20200</math>. Adding in the original center box we have <math> 20200 + 1 = \boxed{\textbf{(C) }20201}</math>. | We can see that each figure <math>n</math> has a central box and 4 columns of <math>n</math> 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 <math>\sum_{n=1}^{100} n = 5050</math> squares. <math>4 \cdot 5050 = 20200</math>. Adding in the original center box we have <math> 20200 + 1 = \boxed{\textbf{(C) }20201}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let <math>a_n</math> be the number of squares in figure <math>n</math>. We can easily see that | ||
+ | <cmath>a_0=4\cdot 0+1</cmath> | ||
+ | <cmath>a_1=4\cdot 1+1</cmath> | ||
+ | <cmath>a_2=4\cdot 3+1</cmath> | ||
+ | <cmath>a_3=4\cdot 6+1.</cmath> | ||
+ | Note that in <math>a_n</math>, the number multiplied by the 4 is the <math>n</math>th triangular number. Hence, <math>a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}</math>. | ||
+ | ~qkddud~ | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>f_n</math> denote the number of unit cubes in a figure. We have | ||
+ | <cmath>f_0=1</cmath> | ||
+ | <cmath>f_1=5</cmath> | ||
+ | <cmath>f_2=13</cmath> | ||
+ | <cmath>f_3=25</cmath> | ||
+ | <cmath>f_4=41</cmath> | ||
+ | <cmath>...</cmath> | ||
+ | |||
+ | Computing the difference between the number of cubes in each figure yields | ||
+ | <cmath>4,8,12,16,...</cmath> | ||
+ | It is easy to notice that this is an arithmetic sequence, with the first term being <math>4</math> and the difference being <math>4</math>. Let this sequence be <math>a_n</math> | ||
+ | |||
+ | From <math>f_0</math> to <math>f_{100}</math>, the sequence will have <math>100</math> terms. Using the arithmetic sum formula yields | ||
+ | |||
+ | <cmath>S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}</cmath> | ||
+ | <cmath>=50(2\cdot 4+99\cdot 4)</cmath> | ||
+ | <cmath>=50(101\cdot 4)</cmath> | ||
+ | <cmath>=200\cdot 101</cmath> | ||
+ | <cmath>=20200</cmath> | ||
+ | |||
+ | So <math>f_{100}=1+20200=\boxed{\textbf{(C) }20201}</math> unit cubes. | ||
+ | |||
+ | ~ljlbox | ||
+ | |||
+ | == Solution 6(Newton's Forward Differences) == | ||
+ | We know that 1 and 5 differ by 4, 5 and 13 differ by 8, and 13 and 25 differ by 12. Hence the differences are 4, 8, and 12, resp. And the differences of the differences area all 4. So by Newton's Forward Difference Formula, we get the 100th figure is(because Figure 0 exists) <math>\dbinom{101-1}{0}+4\dbinom{101-1}{1}+4\dbinom{101-1}{2}=20201</math> or <math>\textbf{(C)}</math> | ||
+ | -vsamc | ||
==See Also== | ==See Also== | ||
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{{AMC12 box|year=2000|num-b=7|num-a=9}} | {{AMC12 box|year=2000|num-b=7|num-a=9}} | ||
{{AMC10 box|year=2000|num-b=11|num-a=13}} | {{AMC10 box|year=2000|num-b=11|num-a=13}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:42, 25 August 2020
- The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.
Contents
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
Solution 1
We can divide up figure to get the sum of the sum of the first odd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :
The sum of the first odd numbers is , so for figure , there are unit squares. We plug in to get .
Solution 2
Using the recursion from solution 1, we see that the first differences of form an arithmetic progression, and consequently that the second differences are constant and all equal to . Thus, the original sequence can be generated from a quadratic function.
If , and , , and , we get a system of three equations in three variables:
gives
gives
gives
Plugging in into the last two equations gives
Dividing the second equation by 2 gives the system:
Subtracting the first equation from the second gives , and hence . Thus, our quadratic function is:
Calculating the answer to our problem, , which is choice .
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
Solution 6(Newton's Forward Differences)
We know that 1 and 5 differ by 4, 5 and 13 differ by 8, and 13 and 25 differ by 12. Hence the differences are 4, 8, and 12, resp. And the differences of the differences area all 4. So by Newton's Forward Difference Formula, we get the 100th figure is(because Figure 0 exists) or -vsamc
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.