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

m (→‎Solution 2)
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<cmath>f(14)=13</cmath>
 
<cmath>f(14)=13</cmath>
  
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f(15)=15
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<cmath>f(15)=15</cmath>
−
.....
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<cmath>.....</cmath>
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Notice how it repeats every 6 times where f(n)=n The number will always be an odd multiple of 3. The pattern of the numbers that follow will always be +3, +2, +0, -1, +0
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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 +3, +2, +0, -1, +0.
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The closest one to 2018 is 2013 so
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The closest odd multiple of <math>3</math> to <math>2018</math> is <math>2013</math>, so we have
 
<cmath>f(2013)=2013</cmath>
 
<cmath>f(2013)=2013</cmath>
 
<cmath>f(2014)=2016</cmath>
 
<cmath>f(2014)=2016</cmath>
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<cmath>f(2016)=2018</cmath>
 
<cmath>f(2016)=2018</cmath>
 
<cmath>f(2017)=2017</cmath>
 
<cmath>f(2017)=2017</cmath>
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<cmath>f(2018)=\boxed{2017}</cmath>$
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<cmath>f(2018)=\boxed{2017}.</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}}
 
{{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}}
 +
{{AMC12 box|year=2018|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:31, 16 February 2018

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

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 closest odd multiple of $3$ to $2018$ is $2013$, so we have \[f(2013)=2013\] \[f(2014)=2016\] \[f(2015)=2018\] \[f(2016)=2018\] \[f(2017)=2017\] \[f(2018)=\boxed{2017}.\]

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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