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

(Solution 3 (Algebra))
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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==
  
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<cmath>f(15)=15</cmath>
 
<cmath>f(15)=15</cmath>
 
<cmath>.....</cmath>
 
<cmath>.....</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>+3</math>, <math>+2</math>, <math>+0</math>, <math>-1</math>, <math>+0</math>.
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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 <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>
 
<cmath>f(2013)=2013</cmath>
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<cmath>f(2017)=2017</cmath>
 
<cmath>f(2017)=2017</cmath>
 
<cmath>f(2018)=\boxed{(B) 2017}.</cmath>
 
<cmath>f(2018)=\boxed{(B) 2017}.</cmath>
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minor edits by bunny1
  
 
==Solution 2 (Bashy Pattern Finding)==
 
==Solution 2 (Bashy Pattern Finding)==
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<math>1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...</math>. Examining, we see that every number <math>x</math> where <math>x \equiv 1\pmod 6</math> has <math>f(x)=x</math>, <math>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)=2017.</math> <math>\boxed B</math>
 
<math>1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...</math>. Examining, we see that every number <math>x</math> where <math>x \equiv 1\pmod 6</math> has <math>f(x)=x</math>, <math>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)=2017.</math> <math>\boxed B</math>
  
==Solution 4 (Algebra)==
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==Solution 3 (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>
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~AopsUser101
 
~AopsUser101
  
==Solution 4 (Algebra)==
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==Video Solution==
Solve the difference equation using the method of undermined coefficients and get
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<cmath> f(n) = \frac{4}{\sqrt 3} \sin\frac{2n\pi}{3} + n + 1</cmath>
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https://www.youtube.com/watch?v=aubDsjVFFTc
Then <math>f(2018) = \frac{4}{\sqrt 3} (- \frac{\sqrt 3}{2}) + 2018 + 1 = 2017</math>.
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~bunny1
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https://youtu.be/ub6CdxulWfc
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 21:16, 14 January 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

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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