Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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==Solution 6== | ==Solution 6== | ||
− | < | + | <cmath>\begin{align*} |
− | + | f(n)&=f(n-1)-f(n-2)+n \\ | |
− | Subtracting those two and rearranging gives < | + | f(n-1)&=f(n-2)-f(n-3)+n-1 |
− | + | \end{align*}</cmath> | |
− | Subtracting those two gives <math>f(n)-3f(n-1)+4f(n-2)-3f(n-3)+f(n-4)=0</math> | + | Subtracting those two and rearranging gives |
− | The characteristic polynomial is <math>x^4-3x^3+4x^2-3x+1=0</math> | + | <cmath>\begin{align*} |
− | <math>x=1</math> is a root, so using synthetic division results in <math>(x-1)(x^3-2x^2+2x-1)=0</math> | + | f(n)-2f(n-1)+2f(n-2)-f(n-3)&=1 \\ |
− | <math>x=1</math> is a root, so using synthetic division results in <math>(x-1)^2(x^2-x+1)=0</math> | + | f(n-1)-2f(n-2)+2f(n-3)-f(n-4)&=1 |
− | <math>x^2-x+1=0</math> has roots <math>x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}</math> | + | \end{align*}</cmath> |
− | And <cmath>f(n)=(An+D)\cdot1^n+B\cdot\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)^n+C\cdot\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^n</cmath> | + | Subtracting those two gives <math>f(n)-3f(n-1)+4f(n-2)-3f(n-3)+f(n-4)=0.</math> |
+ | |||
+ | The characteristic polynomial is <math>x^4-3x^3+4x^2-3x+1=0.</math> | ||
+ | |||
+ | <math>x=1</math> is a root, so using synthetic division results in <math>(x-1)(x^3-2x^2+2x-1)=0.</math> | ||
+ | |||
+ | <math>x=1</math> is a root, so using synthetic division results in <math>(x-1)^2(x^2-x+1)=0.</math> | ||
+ | |||
+ | <math>x^2-x+1=0</math> has roots <math>x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}.</math> | ||
+ | |||
+ | And <cmath>f(n)=(An+D)\cdot1^n+B\cdot\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)^n+C\cdot\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^n.</cmath> | ||
Plugging in <math>n=1</math>, <math>n=2</math>, <math>n=3</math>, and <math>n=4</math> results in a system of <math>4</math> linear equations<math>\newline</math> | Plugging in <math>n=1</math>, <math>n=2</math>, <math>n=3</math>, and <math>n=4</math> results in a system of <math>4</math> linear equations<math>\newline</math> | ||
Solving them gives <math>A=1, \ B=\frac{1}{2}-\frac{i\sqrt{3}}{2}, \ C=\frac{1}{2}+\frac{i\sqrt{3}}{2}, \ D=1.</math> Note that you can guess <math>A=1</math> by answer choices. | Solving them gives <math>A=1, \ B=\frac{1}{2}-\frac{i\sqrt{3}}{2}, \ C=\frac{1}{2}+\frac{i\sqrt{3}}{2}, \ D=1.</math> Note that you can guess <math>A=1</math> by answer choices. | ||
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First, we need to know whether or not <math>2016</math> is part of the skip or repeat. We notice that <math>f(6),f(12), \ldots,f(6n)</math> all satisfy <math>6+6(n-1)=n,</math> and we know that <math>2016</math> satisfies this, leaving <math>n=336.</math> Therefore, we know that <math>2016</math> is part of the repeat section. But what number does it repeat? | First, we need to know whether or not <math>2016</math> is part of the skip or repeat. We notice that <math>f(6),f(12), \ldots,f(6n)</math> all satisfy <math>6+6(n-1)=n,</math> and we know that <math>2016</math> satisfies this, leaving <math>n=336.</math> Therefore, we know that <math>2016</math> is part of the repeat section. But what number does it repeat? | ||
− | We know that the repeat period is <math>2,</math> and it follows that pattern of <math>1,1,8,8,7,7.</math> Again, since <math>f(6) = f(5)</math> and so on for the repeat section, <math>f(2016)=f(2015),</math> so we don't need to worry about which one, since it repeats with period <math>2.</math> We see that the repeat pattern of <math>f(6),f(12),\ldots,f(6n)</math> follows <math>8,14,20,</math> it is an arithmetic sequence with common difference <math>6.</math> Therefore, <math>2016</math> is the <math>335</math>th term of this, but including | + | We know that the repeat period is <math>2,</math> and it follows that pattern of <math>1,1,8,8,7,7.</math> Again, since <math>f(6) = f(5)</math> and so on for the repeat section, <math>f(2016)=f(2015),</math> so we don't need to worry about which one, since it repeats with period <math>2.</math> We see that the repeat pattern of <math>f(6),f(12),\ldots,f(6n)</math> follows <math>8,14,20,</math> it is an arithmetic sequence with common difference <math>6.</math> Therefore, <math>2016</math> is the <math>335</math>th term of this, but including <math>1,</math> it is <math>336\cdot6+1=\boxed{\textbf{(B) } 2017}.</math> |
~CharmaineMa07292010 | ~CharmaineMa07292010 |
Latest revision as of 02:11, 11 September 2023
- 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
![$\begin{cases}\begin{aligned} a_{n+1} &= a_{n} - a_{n-1} + (n+1) \\ a_{n} &= a_{n-1} - a_{n-2} + n \end{aligned}.\end{cases}$](http://latex.artofproblemsolving.com/5/1/0/510802e58f1d2347ac5c91488f8444ecfa90d528.png)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.