2017 AMC 12A Problem 7

Attempt Problem 7 of the 2017 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

$2017$

$2018$

$4034$

$4035$

$4036$

Answer: B

Concepts:recursion pattern recognition

Difficulty rating: 1380

Solution:

Listing values: $f(1)=2,$ $f(2)=f(1)+1=3,$ $f(3)=f(1)+2=4,$ $f(4)=f(3)+1=5,$ suggesting $f(n)=n+1.$

Both rules are consistent with $f(n)=n+1:$ for even $n,$ $(n-1)+1+1=n+1,$ and for odd $n,$ $(n-2)+1+2=n+1.$ Since the recursion determines $f$ uniquely, $f(2017)=2018.$

Thus, the correct answer is B.

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