Difference between revisions of "2018 AMC 10B Problems/Problem 20"

(Solution 3 (Algebra))
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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.

Problem

A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\]for all integers $n \geq 3$. What is $f(2018)$?

$\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}$

Solution 1 (A Bit Bashy)

Start out by listing some terms of the sequence. \[f(1)=1\] \[f(2)=1\]

\[f(3)=3\] \[f(4)=6\] \[f(5)=8\] \[f(6)=8\] \[f(7)=7\] \[f(8)=7\]

\[f(9)=9\] \[f(10)=12\] \[f(11)=14\] \[f(12)=14\] \[f(13)=13\] \[f(14)=13\]

\[f(15)=15\] \[.....\] Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$, and the pattern of numbers that follow will always be $+2$, $+3$, $+2$, $+0$, $-1$, $+0$. The largest odd multiple of $3$ smaller than $2018$ is $2013$, so we have \[f(2013)=2013\] \[f(2014)=2016\] \[f(2015)=2018\] \[f(2016)=2018\] \[f(2017)=2017\] \[f(2018)=\boxed{(B) 2017}.\] minor edits by bunny1

Solution 2 (Bashy Pattern Finding)

Writing out the first few values, we get: $1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...$. Examining, we see that every number $x$ where $x \equiv 1\pmod 6$ has $f(x)=x$, $f(x+1)=f(x)=x$, and $f(x-1)=f(x-2)=x+1$. The greatest number that's $1\pmod{6}$ and less $2018$ is $2017$, so we have $f(2017)=f(2018)=2017.$ $\boxed B$

Solution 3 (Algebra)

\[f(n)=f(n-1)-f(n-2)+n.\] \[f(n-1)=f(n-2)-f(n-3)+n-1.\] Adding the two equations, we have that \[f(n)=2n-1-f(n-3).\] Hence, $f(n)+f(n-3)=2n-1$. After plugging in $n-3$ to the equation above and doing some algebra, we have that $f(n)-f(n-6)=6$. Consequently, \[f(2018)-f(2012)=6.\] \[f(2012)-f(2006)=6.\] \[\ldots\] \[f(8)-f(2)=6.\] Adding these $336$ equations up, we have that $f(2018)-f(2)=6 \cdot 336$ and $f(2018)=\boxed{2017}$.

~AopsUser101

Solution 4 (Easier Pattern Finding)

Note: In this solution, we define the sequence $A$ to satisfy $a_n = f(n),$ where $a_n$ represents the $n$th term of the sequence $A.$

To begin, we consider the sequence $B$ formed when we take the difference of consecutive terms between $A.$ Define $b_n = a_{n+1} - a_n.$ Notice that for $n \ge 4,$ we have \[\begin{cases} a_{n+1} = a_{n} - a_{n-1} + (n+1) \\ a_{n} = a_{n-1} - a_{n-2} + n. \end{cases}\] Notice that subtracting the second equation from the first, we see that $b_{n} = b_{n-1} - b_{n-2} + 1.$

Evaluating the first few terms of $B$, we see that we get \[\underbrace{0, 2, 3, 2, 0, -1,}_{\text{cycle}} 0, 2, 3, 2, 0, -1, \cdots,\] in which we see it has a cycle period of $6$! Notice that any six consecutive terms of $B$ sum to $6,$ after which we see that $a_n = a_{n-6} + 6.$

Therefore, $a_{2018} = a_{2012} + 6 = \cdots = a_{2} + 2016 = \boxed{2017}.$

Note: If you didn’t notice that $B$ repeated directly in the solution above, you could also take the finite differences of the sequence $b_n$ so that you could define $c_n = b_{n+1} - b_n.$ Using a similar method as above through reindexing and then subtracting, you could find that $c_n = c_{n-1} - c_{n-2}.$ Using the fact that $b_1 = 0, \ b_2 = 2, $and $b_3 = 3,$ we see that the sequence $C$ looks like \[2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, \cdots\] However, $C$ has a cycle period of $6$! Notice that any six consecutive terms of $C$ sum to $0,$ which implies that $b_n = b_{n+6}.$

Video Solution

https://www.youtube.com/watch?v=aubDsjVFFTc

~bunny1

https://youtu.be/ub6CdxulWfc

~savannahsolver

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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