Difference between revisions of "2017 AMC 12A Problems/Problem 7"
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This is a recursive function, which means the function is used to evaluate itself. To solve this, we must identify the base case, <math>f(1)=2</math>. We also know that when <math>n</math> is odd, <math>f(n)=f(n-2)+2</math>. Thus we know that <math>f(2017)=f(2015)+2</math>. Thus we know that n will always be odd in the recursion of <math>f(2017)</math>, and we add two each recursive cycle, which there are <math>1008</math> of. Thus the answer is <math>1008*2+2=2018</math>. | This is a recursive function, which means the function is used to evaluate itself. To solve this, we must identify the base case, <math>f(1)=2</math>. We also know that when <math>n</math> is odd, <math>f(n)=f(n-2)+2</math>. Thus we know that <math>f(2017)=f(2015)+2</math>. Thus we know that n will always be odd in the recursion of <math>f(2017)</math>, and we add two each recursive cycle, which there are <math>1008</math> of. Thus the answer is <math>1008*2+2=2018</math>. | ||
<math>\boxed{\textbf{(B)}}</math> | <math>\boxed{\textbf{(B)}}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2017|ab=A|num-b=6|num-a=8}} | ||
Revision as of 15:54, 8 February 2017
Problem
Define a function on the positive integers recursively by 
, 
if 
is even, and 
if is odd and greater than 
. What is 
?
Solution
This is a recursive function, which means the function is used to evaluate itself. To solve this, we must identify the base case, 
. We also know that when is odd, 
. Thus we know that 
. Thus we know that n will always be odd in the recursion of , and we add two each recursive cycle, which there are 
of. Thus the answer is 
.
See Also
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 | |
