2018 AMC 10B Problems/Problem 20
- 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 (Algebra)
For all integers note that It follows that ~MRENTHUSIASM
Solution 2 (Algebra)
For all integers we rearrange the given equation: For all integers it follows that For all integers we add and For all integers it follows that For all integers we subtract from From we have the following system of equations: We add these equations up to get from which
~AopsUser101 ~MRENTHUSIASM
Solution 3 (Finite Differences)
Preamble: In this solution, we define the sequence to satisfy where represents the th term of the sequence This solution will show a few different perspectives. Even though it may not be as quick as some of the solutions above, I feel like it is an interesting concept, and may be more motivated.
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
If you didn’t notice that repeated directly in the solution above, you could also, possibly more naturally, 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 The sum of any six consecutive terms of a sequence which satisfies such a recursion is in which you have that In the case in which finite differences didn’t reduce to such a special recursion, you could still find the first few terms of to see if there are any patterns, now that you have a much simpler sequence. Doing so in this case, it can also be seen by seeing that the sequence looks like in which the same result follows.
Using the fact that repeats every six terms, this motivates us to look at the sequence more carefully. Doing so, we see that looks like (If you tried pattern finding on sequence directly, you could also arrive at this result, although I figured defining a second sequence based on finite differences was more motivated.)
Now, there are two ways to finish.
Finish Method #1: Notice that any six consecutive terms of sum to after which we see that Therefore,
Finish Method #2: Notice that
~Professor-Mom
Solution 4 (Pattern)
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
Solution 5 (Pattern)
Writing out the first few values, we get We see that every number where has and The greatest number that's and less than is so we have
Solution 6
Subtracting those two and rearranging gives Subtracting those two gives
The characteristic polynomial is
is a root, so using synthetic division results in
is a root, so using synthetic division results in
has roots
And Plugging in , , , and results in a system of linear equations Solving them gives Note that you can guess by answer choices.
So plugging in results in ~ryanbear
Solution 7
We utilize patterns to solve this equation: We realize that the pattern repeats itself. For every six terms, there will be four terms that we repeat, and two terms that we don't repeat. We will exclude the first two for now, because they don't follow this pattern.
First, we need to know whether or not is part of the skip or repeat. We notice that all satisfy and we know that satisfies this, leaving Therefore, we know that is part of the repeat section. But what number does it repeat?
We know that the repeat period is and it follows that pattern of Again, since and so on for the repeat section, so we don't need to worry about which one, since it repeats with period We see that the repeat pattern of follows it is an arithmetic sequence with common difference Therefore, is the th term of this, but including it is
~CharmaineMa07292010
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 |
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