Difference between revisions of "2018 AMC 10B Problems/Problem 20"
(→Solution 2) |
|||
| Line 5: | Line 5: | ||
<math>\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}</math> | <math>\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
<math>f\left(n\right) = f\left(n - 1\right) - f\left(n - 2\right) + n</math> | <math>f\left(n\right) = f\left(n - 1\right) - f\left(n - 2\right) + n</math> | ||
| Line 18: | Line 18: | ||
Thus, <math>f\left(2018\right) = 2016 + f\left(2\right) = 2017</math>. <math>\boxed{B}</math> | Thus, <math>f\left(2018\right) = 2016 + f\left(2\right) = 2017</math>. <math>\boxed{B}</math> | ||
| + | ==Solution 2== | ||
| + | Start out by listing some terms of the sequence. | ||
| + | <cmath>f(1)=1</cmath> | ||
| + | <cmath>f(2)=1</cmath> | ||
| + | |||
| + | <cmath>f(3)=3</cmath> | ||
| + | <cmath>f(4)=6</cmath> | ||
| + | <cmath>f(5)=8</cmath> | ||
| + | <cmath>f(6)=8</cmath> | ||
| + | <cmath>f(7)=7</cmath> | ||
| + | <cmath>f(8)=7</cmath> | ||
| + | |||
| + | <cmath>f(9)=9</cmath> | ||
| + | <cmath>f(10)=12</cmath> | ||
| + | <cmath>f(11)=14</cmath> | ||
| + | <cmath>f(12)=14</cmath> | ||
| + | <cmath>f(13)=13</cmath> | ||
| + | <cmath>f(14)=13</cmath> | ||
| + | |||
| + | f(15)=15 | ||
| + | ..... | ||
| + | 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 | ||
| + | The closest one to 2018 is 2013 so | ||
| + | <cmath>f(2013)=2013</cmath> | ||
| + | <cmath>f(2014)=2016</cmath> | ||
| + | <cmath>f(2015)=2018</cmath> | ||
| + | <cmath>f(2016)=2018</cmath> | ||
| + | <cmath>f(2017)=2017</cmath> | ||
| + | <math></math>f(2018)=<math>\boxed{2017}</math><math></math> | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:37, 16 February 2018
Contents
Problem
A function 
is defined recursively by 
and ![\[f(n)=f(n-1)-f(n-2)+n\]](http://latex.artofproblemsolving.com/d/0/8/d088e673fb4436308256a5007dc2aefa5284ca3f.png)
for all integers 
. What is 
?
Solution 1
Thus, 
. 
Solution 2
Start out by listing some terms of the sequence.
![\[f(1)=1\]](http://latex.artofproblemsolving.com/1/3/5/13509491664ca01661b2e3df0cf2d2426cf96a8d.png)
![\[f(2)=1\]](http://latex.artofproblemsolving.com/3/8/7/3877dbfb29493076f321d82923f38b70e84a16f3.png)
![\[f(3)=3\]](http://latex.artofproblemsolving.com/7/d/9/7d9dbc57bbc6f0686719d8e005f822f543db6138.png)
![\[f(4)=6\]](http://latex.artofproblemsolving.com/c/7/4/c74fe089e77ab8f582c978bbdafe77d94f549d89.png)
![\[f(5)=8\]](http://latex.artofproblemsolving.com/d/c/c/dcc2578499cf6285453e307a41c91fad3ff3b079.png)
![\[f(6)=8\]](http://latex.artofproblemsolving.com/a/9/2/a928f250bb4e70fb1fe33d187cc761fe509075f9.png)
![\[f(7)=7\]](http://latex.artofproblemsolving.com/f/8/b/f8b063e8de9523ed80aeca32af20d6e6b6a7dcc6.png)
![\[f(8)=7\]](http://latex.artofproblemsolving.com/c/9/d/c9d162ad7dbe84d72536af79a6172c0a2e3bfad9.png)
![\[f(9)=9\]](http://latex.artofproblemsolving.com/4/0/b/40bfa5563f924853d4037fa1226bf796391fee85.png)
![\[f(10)=12\]](http://latex.artofproblemsolving.com/e/b/d/ebdafbb236c26fa7e298338fa53068a41f145ae6.png)
![\[f(11)=14\]](http://latex.artofproblemsolving.com/e/1/2/e12a24aee66a5370022d262261f0457325d694bd.png)
![\[f(12)=14\]](http://latex.artofproblemsolving.com/d/6/8/d6897b1a6bf3cfb7af408a74def16b767581ae23.png)
![\[f(13)=13\]](http://latex.artofproblemsolving.com/2/2/d/22d51baf70d74b4f26d0db633b756aa2d63bcbcb.png)
![\[f(14)=13\]](http://latex.artofproblemsolving.com/f/3/d/f3d1003bd5de64b3df8fe6611b1d46449047dad2.png)
f(15)=15
.....
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
The closest one to 2018 is 2013 so
![\[f(2013)=2013\]](http://latex.artofproblemsolving.com/3/7/d/37d875556f55b97668a24ff31e104cc3133c0fd4.png)
![\[f(2014)=2016\]](http://latex.artofproblemsolving.com/b/9/8/b98afb766ba334c9923cc0e0a9ce53c4a8963d80.png)
![\[f(2015)=2018\]](http://latex.artofproblemsolving.com/f/3/4/f34c23165106532bd8492882e69350553e7bf8a7.png)
![\[f(2016)=2018\]](http://latex.artofproblemsolving.com/7/f/1/7f1d6733c29ab5566b0f217ca81b205713971ed7.png)
![\[f(2017)=2017\]](http://latex.artofproblemsolving.com/5/c/9/5c9a30a9ec6ca0e9b5919fa5975cddf40ecc816c.png)
$$ (Error compiling LaTeX. Unknown error_msg)f(2018)=
$$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
