Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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~AopsUser101 | ~AopsUser101 | ||
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+ | ==Solution 4 (Easier Pattern Finding)== | ||
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+ | Note: In this solution, we define the sequence <math>A</math> to satisfy <math>a_n = f(n),</math> where <math>a_n</math> represents the <math>n</math>th term of the sequence <math>A.</math> | ||
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+ | To begin, we consider the sequence <math>B</math> formed when we take the difference of consecutive terms between <math>A.</math> Define <math>b_n = a_{n+1} - a_n.</math> Notice that for <math>n \ge 4,</math> we have <cmath>\begin{cases} a_{n+1} = a_{n} - a_{n-1} + (n+1) \\ a_{n} = a_{n-1} - a_{n-2} + n. \end{cases}</cmath> Notice that subtracting the second equation from the first, we see that <math>b_{n} = b_{n-1} - b_{n-2} + 1.</math> | ||
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+ | Evaluating the first few terms of <math>B</math>, we see that we get <cmath>\underbrace{0, 2, 3, 2, 0, -1,}_{\text{cycle}} 0, 2, 3, 2, 0, -1, \cdots, </cmath> in which we see it has a cycle period of <math>6</math>! Notice that any six consecutive terms of <math>B</math> sum to <math>6,</math> after which we see that <math>a_n = a_{n-6} + 6.</math> | ||
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+ | Therefore, <math>a_{2018} = a_{2012} + 6 = \cdots = a_{2} + 2016 = \boxed{2017}.</math> | ||
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+ | Note: If you didn’t notice that <math>B</math> repeated directly in the solution above, you could also take the finite differences of the sequence <math>b_n</math> so that you could define <math>c_n = b_{n+1} - b_n.</math> Using a similar method as above through reindexing and then subtracting, you could find that <math>c_n = c_{n-1} - c_{n-2}.</math> Using the fact that <math>b_1 = 0, \ b_2 = 2, \ </math>and <math>b_3 = 3,</math> we see that the sequence <math>C</math> looks like <cmath>2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, \cdots</cmath> However, <math>C</math> has a cycle period of <math>6</math>! Notice that any six consecutive terms of <math>C</math> sum to <math>0,</math> which implies that <math>b_n = b_{n+6}.</math> | ||
==Video Solution== | ==Video Solution== |
Revision as of 00:43, 24 September 2021
- The following problem is from both the 2018 AMC 12B #18 and 2018 AMC 10B #20, so both problems redirect to this page.
Contents
Problem
A function is defined recursively by and for all integers . What is ?
Solution 1 (A Bit Bashy)
Start out by listing some terms of the sequence.
Notice that whenever is an odd multiple of , and the pattern of numbers that follow will always be , , , , , . The largest odd multiple of smaller than is , so we have minor edits by bunny1
Solution 2 (Bashy Pattern Finding)
Writing out the first few values, we get: . Examining, we see that every number where has , , and . The greatest number that's and less is , so we have
Solution 3 (Algebra)
Adding the two equations, we have that Hence, . After plugging in to the equation above and doing some algebra, we have that . Consequently, Adding these equations up, we have that and .
~AopsUser101
Solution 4 (Easier Pattern Finding)
Note: In this solution, we define the sequence to satisfy where represents the th term of the sequence
To begin, we consider the sequence formed when we take the difference of consecutive terms between Define Notice that for we have Notice that subtracting the second equation from the first, we see that
Evaluating the first few terms of , we see that we get in which we see it has a cycle period of ! Notice that any six consecutive terms of sum to after which we see that
Therefore,
Note: If you didn’t notice that repeated directly in the solution above, you could also take the finite differences of the sequence so that you could define Using a similar method as above through reindexing and then subtracting, you could find that Using the fact that and we see that the sequence looks like However, has a cycle period of ! Notice that any six consecutive terms of sum to which implies that
Video Solution
https://www.youtube.com/watch?v=aubDsjVFFTc
~bunny1
~savannahsolver
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.