2017 AMC 12A Problems/Problem 7

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 two each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$ $\boxed{\textbf{(B)}}$

Page

Toolbox

Search

2017 AMC 12A Problems/Problem 7

Problem

Solution