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

(Solution 3 (Bashy Pattern Finding))
(Solution 4 (Algebra))
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==Solution 4 (Algebra)==
 
==Solution 4 (Algebra)==
<cmath>f(n)=f(n-1)-f(n-2)+n</cmath>
+
<cmath>f(n)=f(n-1)-f(n-2)+n.</cmath>
<cmath>f(n-1)=f(n-2)-f(n-3)+n-1</cmath>
+
<cmath>f(n-1)=f(n-2)-f(n-3)+n-1.</cmath>
Adding the two equations, we have that <cmath>f(n)=2n-1-f(n-3)</cmath>.
+
Adding the two equations, we have that <cmath>f(n)=2n-1-f(n-3).</cmath>  
 
Hence, <math>f(n)+f(n-3)=2n-1</math>.
 
Hence, <math>f(n)+f(n-3)=2n-1</math>.
 
After plugging in <math>n-3</math> to the equation above and doing some algebra, we have that <math>f(n)-f(n-6)=6</math>.
 
After plugging in <math>n-3</math> to the equation above and doing some algebra, we have that <math>f(n)-f(n-6)=6</math>.
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<cmath>f(2012)-f(2006)=6</cmath>
 
<cmath>f(2012)-f(2006)=6</cmath>
 
<cmath>\ldots</cmath>
 
<cmath>\ldots</cmath>
<math>f(8)-f(2)=6</math>
+
<cmath>f(8)-f(2)=6</cmath>
Adding these <math>336</math> equations up, we have that <math>f(2018)-f(2)=6*336</math> and <math>f(2018)=\boxed{2017}</math>
+
Adding these <math>336</math> equations up, we have that <math>f(2018)-f(2)=6*336</math> and <math>f(2018)=\boxed{2017}</math>.
  
 
~AopsUser101
 
~AopsUser101

Revision as of 22:21, 18 July 2019

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

$f\left(n\right) = f\left(n - 1\right) - f\left(n - 2\right) + n$

$= \left(f\left(n - 2\right) - f\left(n - 3\right) + n - 1\right) - f\left(n - 2\right) + n$

$= 2n - 1 - f\left(n - 3\right)$

$= 2n - 1 - \left(2\left(n - 3\right) - 1 - f\left(n - 6\right)\right)$

$= f\left(n - 6\right) + 6$

Thus, $f\left(2018\right) = 2016 + f\left(2\right) = 2017$. $\boxed{B}$

Solution 2 (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 +3, +2, +0, -1, +0. The largest odd multiple of $3$ smaller than $2018$ is $2013$, so we have \[f(2013)=2013\] \[f(2014)=2016\] \[f(2015)=2018\] \[f(2016)=2018\] \[f(2017)=2017\] \[f(2018)=\boxed{(B) 2017}.\]

Solution 3 (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 (mod 6) and less $2018$ is $2017$, so we have $f(2017)=f(2018)=2017.$ $\boxed B$


Solution 4 (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 $f(n)-f(n-6)=6$. We have that: \[f(2018)-f(2012)=6\] \[f(2012)-f(2006)=6\] \[\ldots\] \[f(8)-f(2)=6\] Adding these $336$ equations up, we have that $f(2018)-f(2)=6*336$ and $f(2018)=\boxed{2017}$.

~AopsUser101

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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