Difference between revisions of "2018 AMC 10B Problems/Problem 20"
m (→Solution 2 (A Bit Bashy)) |
|||
(27 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 1: | ||
+ | {{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #18]] and [[2018 AMC 10B Problems|2018 AMC 10B #20]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
Line 5: | Line 7: | ||
<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 1 | + | ==Solution 1 (A Bit Bashy)== |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
Start out by listing some terms of the sequence. | Start out by listing some terms of the sequence. | ||
<cmath>f(1)=1</cmath> | <cmath>f(1)=1</cmath> | ||
Line 39: | Line 28: | ||
<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 +3, +2, +0, -1, +0. | + | 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 | + | 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> | ||
<cmath>f(2014)=2016</cmath> | <cmath>f(2014)=2016</cmath> | ||
Line 47: | Line 36: | ||
<cmath>f(2017)=2017</cmath> | <cmath>f(2017)=2017</cmath> | ||
<cmath>f(2018)=\boxed{(B) 2017}.</cmath> | <cmath>f(2018)=\boxed{(B) 2017}.</cmath> | ||
+ | minor edits by bunny1 | ||
− | ==Solution | + | ==Solution 2 (Bashy Pattern Finding)== |
Writing out the first few values, we get: | Writing out the first few values, we get: | ||
− | <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 1 | + | <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 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> | ||
+ | Adding the two equations, we have that <cmath>f(n)=2n-1-f(n-3).</cmath> | ||
+ | Hence, <math>f(n)+f(n-3)=2n-1</math>. | ||
+ | After plugging in <math>n-3</math> to the equation above and doing some algebra, we have that <math>f(n)-f(n-6)=6</math>. | ||
+ | Consequently, | ||
+ | <cmath>f(2018)-f(2012)=6.</cmath> | ||
+ | <cmath>f(2012)-f(2006)=6.</cmath> | ||
+ | <cmath>\ldots</cmath> | ||
+ | <cmath>f(8)-f(2)=6.</cmath> | ||
+ | Adding these <math>336</math> equations up, we have that <math>f(2018)-f(2)=6 \cdot 336</math> and <math>f(2018)=\boxed{2017}</math>. | ||
+ | |||
+ | ~AopsUser101 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=aubDsjVFFTc | ||
+ | |||
+ | ~bunny1 | ||
+ | |||
+ | https://youtu.be/ub6CdxulWfc | ||
+ | |||
+ | ~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.
Contents
Problem
A function is defined recursively by and for all integers . What is ?
Solution 1 (A Bit Bashy)
Start out by listing some terms of the sequence.
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 minor edits by bunny1
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 3 (Algebra)
Adding the two equations, we have that Hence, . After plugging in to the equation above and doing some algebra, we have that . Consequently, Adding these equations up, we have that and .
~AopsUser101
Video Solution
https://www.youtube.com/watch?v=aubDsjVFFTc
~bunny1
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.