Difference between revisions of "2017 AMC 12A Problems/Problem 7"
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<math> \textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036 </math> | <math> \textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036 </math> | ||
==Solution== | ==Solution== | ||
| − | 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> | ||
Revision as of 16:04, 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 
.
