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 01:43, 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.