Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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+ | {{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #18]] and [[2018 AMC 10B Problems|2018 AMC 10B #20]]}} | ||
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==Problem== | ==Problem== | ||
A function <math>f</math> is defined recursively by <math>f(1)=f(2)=1</math> and <cmath>f(n)=f(n-1)-f(n-2)+n</cmath>for all integers <math>n \geq 3</math>. What is <math>f(2018)</math>? | A function <math>f</math> is defined recursively by <math>f(1)=f(2)=1</math> and <cmath>f(n)=f(n-1)-f(n-2)+n</cmath>for all integers <math>n \geq 3</math>. What is <math>f(2018)</math>? | ||
− | <math>\textbf{(A)} \ | + | <math>\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020</math> |
+ | |||
+ | ==Solution 1 (Algebra)== | ||
+ | For all integers <math>n \geq 7,</math> note that | ||
+ | <cmath>\begin{align*} | ||
+ | f(n)&=f(n-1)-f(n-2)+n \\ | ||
+ | &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\ | ||
+ | &=-f(n-3)+2n-1 \\ | ||
+ | &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\ | ||
+ | &=-f(n-4)+f(n-5)+n+2 \\ | ||
+ | &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\ | ||
+ | &=f(n-6)+6. | ||
+ | \end{align*}</cmath> | ||
+ | It follows that | ||
+ | <cmath>\begin{align*} | ||
+ | f(2018)&=f(2012)+6 \\ | ||
+ | &=f(2006)+12 \\ | ||
+ | &=f(2000)+18 \\ | ||
+ | & \ \vdots \\ | ||
+ | &=f(2)+2016 \\ | ||
+ | &=\boxed{\textbf{(B) } 2017}. | ||
+ | \end{align*}</cmath> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Algebra)== | ||
+ | For all integers <math>n\geq3,</math> we rearrange the given equation: <cmath>f(n)-f(n-1)+f(n-2)=n. \hspace{28.25mm}(1)</cmath> | ||
+ | For all integers <math>n\geq4,</math> it follows that <cmath>f(n-1)-f(n-2)+f(n-3)=n-1. \hspace{15mm}(2)</cmath> | ||
+ | For all integers <math>n\geq4,</math> we add <math>(1)</math> and <math>(2):</math> <cmath>f(n)+f(n-3)=2n-1. \hspace{38.625mm}(3)</cmath> | ||
+ | For all integers <math>n\geq7,</math> it follows that <cmath>f(n-3)+f(n-6)=2n-7. \hspace{32mm}(4)</cmath> | ||
+ | For all integers <math>n\geq7,</math> we subtract <math>(4)</math> from <math>(3):</math> <cmath>f(n)-f(n-6)=6. \hspace{47.5mm}(5)</cmath> | ||
+ | From <math>(5),</math> we have the following system of <math>336</math> equations: | ||
+ | <cmath>\begin{align*} | ||
+ | f(2018)-f(2012)&=6, \\ | ||
+ | f(2012)-f(2006)&=6, \\ | ||
+ | f(2006)-f(2000)&=6, \\ | ||
+ | & \ \vdots \\ | ||
+ | f(8)-f(2)&=6. | ||
+ | \end{align*}</cmath> | ||
+ | We add these equations up to get <cmath>f(2018)-f(2)=6\cdot336=2016,</cmath> from which <math>f(2018)=f(2)+2016=\boxed{\textbf{(B) } 2017}.</math> | ||
+ | |||
+ | ~AopsUser101 ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 3 (Finite Differences)== | ||
+ | |||
+ | Preamble: 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> 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 <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 <center><math>\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}</math></center> Notice that subtracting the second equation from the first, we see that <math>b_{n} = b_{n-1} - b_{n-2} + 1.</math> | ||
+ | |||
+ | If you didn’t notice that <math>B</math> repeated directly in the solution above, you could also, possibly more naturally, 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> The sum of any six consecutive terms of a sequence which satisfies such a recursion is <math>0,</math> in which you have that <math>b_{n} = b_{n+6}.</math> In the case in which finite differences didn’t reduce to such a special recursion, you could still find the first few terms of <math>C</math> 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 <math>C</math> looks like <cmath>\underbrace{2, 1, -1, -2, -1, 1,}_{\text{cycle period}} 2, 1, -1, -2, -1, 1, \ldots</cmath> in which the same result follows. | ||
+ | |||
+ | Using the fact that <math>B</math> repeats every six terms, this motivates us to look at the sequence <math>B</math> more carefully. Doing so, we see that <math>B</math> looks like <cmath>\underbrace{2, 3, 2, 0, -1, 0,}_{\text{cycle period}} 2, 3, 2, 0, -1, 0, \ldots</cmath> (If you tried pattern finding on sequence <math>B</math> 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 <math>B</math> sum to <math>6,</math> after which we see that <math>a_n = a_{n-6} + 6.</math> Therefore, <math>a_{2018} = a_{2012} + 6 = \cdots = a_{2} + 2016 = \boxed{\textbf{(B) } 2017}.</math> | |
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− | < | ||
− | < | ||
− | < | + | Finish Method #2: Notice that <math>a_{2018} = a_{2017} - a_{2016} + 2018 = B_{2016} + 2018 = -1 + 2018 = \boxed{\textbf{(B) } 2017}.</math> |
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− | + | ~Professor-Mom | |
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− | <cmath>f(15)=15 | + | ==Solution 4 (Bash)== |
− | + | Start out by listing some terms of the sequence. | |
− | Notice that <math>f(n)=n</math> whenever <math>n</math> is an odd multiple of <math>3</math>, and the pattern of numbers that follow will always be <math>+3</math>, <math>+2</math>, <math>+0</math>, <math>-1</math>, <math>+0</math>. | + | <cmath>\begin{align*} |
+ | f(1)&=1 \\ | ||
+ | f(2)&=1 \\ | ||
+ | f(3)&=3 \\ | ||
+ | f(4)&=6 \\ | ||
+ | f(5)&=8 \\ | ||
+ | f(6)&=8 \\ | ||
+ | f(7)&=7 \\ | ||
+ | f(8)&=7 \\ | ||
+ | f(9)&=9 \\ | ||
+ | f(10)&=12 \\ | ||
+ | f(11)&=14 \\ | ||
+ | f(12)&=14 \\ | ||
+ | f(13)&=13 \\ | ||
+ | f(14)&=13 \\ | ||
+ | f(15)&=15 \\ | ||
+ | & \ \vdots | ||
+ | \end{align*}</cmath> | ||
+ | Notice that <math>f(n)=n</math> whenever <math>n</math> is an odd multiple of <math>3</math>, and the pattern of numbers that follow will always be <math>+2</math>, <math>+3</math>, <math>+2</math>, <math>+0</math>, <math>-1</math>, <math>+0</math>. | ||
The largest odd multiple of <math>3</math> smaller than <math>2018</math> is <math>2013</math>, so we have | The largest odd multiple of <math>3</math> smaller than <math>2018</math> is <math>2013</math>, so we have | ||
− | <cmath>f(2013)=2013 | + | <cmath>\begin{align*} |
− | + | f(2013)&=2013 \\ | |
− | + | f(2014)&=2016 \\ | |
− | <cmath>f( | + | f(2015)&=2018 \\ |
− | < | + | f(2016)&=2018 \\ |
− | < | + | f(2017)&=2017 \\ |
+ | f(2018)&=\boxed{\textbf{(B) } 2017}. | ||
+ | \end{align*}</cmath> | ||
+ | minor edits by bunny1 | ||
+ | |||
+ | ==Solution 5 (Bash)== | ||
+ | Writing out the first few values, we get | ||
+ | <cmath>1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19,\ldots.</cmath> We see that every number <math>x</math> where <math>x \equiv 1\pmod 6</math> has <math>f(x)=x,f(x+1)=f(x)=x,</math> and <math>f(x-1)=f(x-2)=x+1.</math> The greatest number that's <math>1\pmod{6}</math> and less <math>2018</math> is <math>2017,</math> so we have <math>f(2017)=f(2018)=\boxed{\textbf{(B) } 2017}.</math> | ||
+ | |||
+ | ==Video Solution== | ||
− | + | https://www.youtube.com/watch?v=aubDsjVFFTc | |
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− | + | ~bunny1 | |
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− | + | https://youtu.be/ub6CdxulWfc | |
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 09:16, 14 October 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 (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 (Bash)
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 5 (Bash)
Writing out the first few values, we get We see that every number where has and The greatest number that's and less is so we have
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.