Difference between revisions of "2018 AMC 10B Problems/Problem 20"
Line 28: | 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 <math>+3</math>, <math>+2</math>, <math>+0</math>, <math>-1</math>, <math>+0</math>. | + | 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 largest odd multiple of <math>3</math> smaller than <math>2018</math> is <math>2013</math>, so we have | 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> |
Revision as of 15:08, 1 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
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
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.