Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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==Solution 3(Bashy Pattern Finding)== | ==Solution 3(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 (mod 6) and less <math>2018</math> is <math>2017</math>, so we have <math>f(2017)=f(2018)=2017.</math> \boxed B | + | <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> |
− | + | ||
==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}} | ||
{{AMC12 box|year=2018|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2018|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:53, 19 February 2018
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 closest odd multiple of to 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
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.