2017 AMC 12A Problems/Problem 7

Revision as of 12:49, 9 February 2017 by EricF (talk | contribs) (Solution)

Problem

Define a function on the positive integers recursively by $f(1) = 2$, $f(n) = f(n-1) + 2$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than $1$. What is $f(2017)$?

$\textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036$

Solution

This is a recursive function, which means the function is used to evaluate itself. To solve this, we must identify the base case, $f(1)=2$. We also know that when $n$ is odd, $f(n)=f(n-2)+2$. Thus we know that $f(2017)=f(2015)+2$. Thus we know that n will always be odd in the recursion of $f(2017)$, and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$. which is answer $\boxed{\textbf{(B)}}$

Alternatively, simply download java and run this method and input $n=2017$:

   public static int f(int n){
       if(n == 1)
           return 2;
       else if(n % 2 == 0)
           return f(n-1) + 2;
       else
           return f(n-2) + 2;
   }

It will return $f(2017) = 2018$ which is answer choice $\boxed{\textbf{(B)}}$

See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
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