Difference between revisions of "2017 AMC 12A Problems/Problem 7"

Revision as of 17:17, 8 February 2017

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 $1008*2+2=2018$ . $\boxed{\textbf{(B)}}$

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
 {{AMC12 box|year=2017|ab=A|num-b=6|num-a=8}}
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Difference between revisions of "2017 AMC 12A Problems/Problem 7"

Revision as of 17:17, 8 February 2017

Problem

Solution

See Also