Difference between revisions of "2000 AMC 12 Problems/Problem 8"
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==Solution 5== | ==Solution 5== | ||
− | Let <math>a_n</math> be the number of squares in figure n. We can easily see that | + | 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_0=4\cdot 0+1</cmath> | ||
<cmath>a_1=4\cdot 1+1</cmath> | <cmath>a_1=4\cdot 1+1</cmath> | ||
<cmath>a_2=4\cdot 3+1</cmath> | <cmath>a_2=4\cdot 3+1</cmath> | ||
<cmath>a_3=4\cdot 6+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> | + | 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=20201,</math> which is <math>\boxed{\text{C}}</math>. |
==See Also== | ==See Also== |
Revision as of 21:21, 14 February 2019
- 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?
Solution
Solution 1
We have a recursion:
.
We add increasing multiples of each time we go up a figure. So, for example, in figure
, we would have
added on to Figure 0, which is
. This yields
So, to go from Figure 0 to 100, we add
.
We then add to the number of squares in Figure 0 to get
, which is choice
Solution 2
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
, which is choice
Solution 3
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 4
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
which is answer choice
Solution 5
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,
which is
.
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.