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

Revision as of 20:03, 1 August 2020

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 $+3$ , $+2$ , $+0$ , $-1$ , $+0$ . 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 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

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.

@@ Line 39: / Line 39: @@
 <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)==
+==Solution 3 (Algebra)==
 <cmath>f(n)=f(n-1)-f(n-2)+n.</cmath>
 <cmath>f(n-1)=f(n-2)-f(n-3)+n-1.</cmath>

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