Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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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 (mod 6) 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 1 (mod 6) 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)== | ||
+ | <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>. | ||
+ | We have that: | ||
+ | <cmath>f(2018)-f(2012)=6</cmath> | ||
+ | <cmath>f(2012)-f(2006)=6</cmath> | ||
+ | <cmath>\ldots</cmath> | ||
+ | <math>f(8)-f(2)=6</math> | ||
+ | Adding these <math>336</math> equations up, we have that <math>f(2018)-f(2)=6*336</math> and <math>f(2018)=\boxed{2017}</math> | ||
+ | |||
+ | ~AopsUser101 | ||
==See Also== | ==See Also== |
Revision as of 21:21, 18 July 2019
Contents
Problem
A function is defined recursively by and for all integers . What is ?
Solution 1
Thus, .
Solution 2 (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 +3, +2, +0, -1, +0. The largest odd multiple of smaller than is , so we have
Solution 3 (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 1 (mod 6) and less is , so we have
Solution 4 (Algebra)
Adding the two equations, we have that . Hence, . After plugging in to the equation above and doing some algebra, we have that . We have that: Adding these equations up, we have that and
~AopsUser101
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.