Difference between revisions of "2000 AMC 12 Problems/Problem 8"
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<math>4 \cdot 1+4 \cdot 2+...+4 \cdot 99+4\cdot 100=4 \cdot 5050=20200</math>. | <math>4 \cdot 1+4 \cdot 2+...+4 \cdot 99+4\cdot 100=4 \cdot 5050=20200</math>. | ||
− | <math>20201</math> | + | We then add <math>20200</math> to the number of squares in Figure 0 to get <math>20201</math>, which is choice <math>\boxed{\text{C}}</math> |
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− | <math>\boxed{\text{C}}</math> | ||
===Solution 2=== | ===Solution 2=== |
Revision as of 09:22, 20 August 2014
- The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.
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:
.
I.E. we add increasing multiples of each time we go up a figure.
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
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.