2018 AMC 10B Problems/Problem 20: Difference between revisions
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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 | ==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> | ||
| Line 53: | Line 53: | ||
~AopsUser101 | ~AopsUser101 | ||
==Solution 4 (Algebra)== | |||
Solve the difference equation using the method of undermined coefficients and get | |||
<cmath> f(n) = \frac{4}{\sqrt 3} \sin\frac{2n\pi}{3} + n + 1</cmath> | |||
Then <math>f(2018) = \frac{4}{\sqrt 3} (- \frac{\sqrt 3}{2}) + 2018 + 1 = 2017</math>. | |||
==See Also== | ==See Also== | ||
Revision as of 20:01, 1 August 2020
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 (A Bit Bashy)
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)
![\[f(15)=15\]](http://latex.artofproblemsolving.com/2/1/2/212c828a007245db77ca3971b3aaf05e71379293.png)
![\[.....\]](http://latex.artofproblemsolving.com/7/f/1/7f14cc043d2c7387fd38f00f4c206086c58ad029.png)
Notice that 
whenever 
is an odd multiple of 
, and the pattern of numbers that follow will always be 
, 
, 
, 
, .
The largest odd multiple of smaller than 
is 
, so we have
![\[f(2013)=2013\]](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)
![\[f(2018)=\boxed{(B) 2017}.\]](http://latex.artofproblemsolving.com/9/1/9/9195b90cb3126903491b021870a941373a297bde.png)
Solution 2 (Bashy Pattern Finding)
Writing out the first few values, we get:
. Examining, we see that every number 
where 
has 
, 
, and 
. The greatest number that's 
and less is 
, so we have 
Solution 4 (Algebra)
![\[f(n)=f(n-1)-f(n-2)+n.\]](http://latex.artofproblemsolving.com/9/a/9/9a94ac69f8fbed96bc5b8f6afff79f9df46bb62e.png)
![\[f(n-1)=f(n-2)-f(n-3)+n-1.\]](http://latex.artofproblemsolving.com/a/8/e/a8efed121c26a52a633ad37400f82159df8088e2.png)
Adding the two equations, we have that ![\[f(n)=2n-1-f(n-3).\]](http://latex.artofproblemsolving.com/c/8/2/c824c7da434b8663b0e63b2c15da08ab6edc95cb.png)
Hence, 
.
After plugging in 
to the equation above and doing some algebra, we have that 
.
Consequently,
![\[f(2018)-f(2012)=6.\]](http://latex.artofproblemsolving.com/f/7/5/f75a27e5bb51d6813c9c9aff0941b9fa6c78703d.png)
![\[f(2012)-f(2006)=6.\]](http://latex.artofproblemsolving.com/0/0/f/00f1ee35cb99659d9e6dacf6b88eb56979e1b8b4.png)
![\[\ldots\]](http://latex.artofproblemsolving.com/2/1/b/21b9e28878899429e6c85bf3a0ba9d65c6a5dd91.png)
![\[f(8)-f(2)=6.\]](http://latex.artofproblemsolving.com/a/2/2/a2261fc915cd0eab5369414acc5a8885d00d5c1d.png)
Adding these 
equations up, we have that 
and 
.
~AopsUser101
Solution 4 (Algebra)
Solve the difference equation using the method of undermined coefficients and get
![\[f(n) = \frac{4}{\sqrt 3} \sin\frac{2n\pi}{3} + n + 1\]](http://latex.artofproblemsolving.com/3/d/d/3dd641f748a4d8ea77d543731390144d71560cda.png)
Then 
.
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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