2017 AMC 12A Problem 7

Problem 7 of 25EasierAlgebra

Define a function on the positive integers recursively by

f(1)=2,

f(n)=f(n-1)+1

n

is even, and

f(n)=f(n-2)+2

n

is odd and greater than

1.

What is

f(2017)?

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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.