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

m (Solution)
m (Solution)
(13 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
Define a function on the positive integers recursively by <math>f(1) = 2</math>, <math>f(n) = f(n-1) + 2</math> if <math>n</math> is even, and <math>f(n) = f(n-2) + 2</math> if <math>n</math> is odd and greater than <math>1</math>. What is <math>f(2017)</math>?
+
Define a function on the positive integers recursively by <math>f(1) = 2</math>, <math>f(n) = f(n-1) + 1</math> if <math>n</math> is even, and <math>f(n) = f(n-2) + 2</math> if <math>n</math> is odd and greater than <math>1</math>. What is <math>f(2017)</math>?
  
 
<math> \textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036 </math>
 
<math> \textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036 </math>
 
==Solution==
 
==Solution==
This is a recursive function, which means the function is used to evaluate itself. To solve this, we must identify the base case, <math>f(1)=2</math>. We also know that when <math>n</math> is odd, <math>f(n)=f(n-2)+2</math>. Thus we know that <math>f(2017)=f(2015)+2</math>. Thus we know that n will always be odd in the recursion of <math>f(2017)</math>, and we add two each recursive cycle, which there are <math>1008</math> of. Thus the answer is <math>1008*2+2=2018</math>.
+
This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, <math>f(1)=2</math>. We also know that when <math>n</math> is odd, <math>f(n)=f(n-2)+2</math>. Thus we know that <math>f(2017)=f(2015)+2</math>. Thus we know that n will always be odd in the recursion of <math>f(2017)</math>, and we add <math>2</math> each recursive cycle, which there are <math>1008</math> of. Thus the answer is <math>1008*2+2=2018</math>, which is answer
<math>\boxed{\textbf{(B)}}</math>
+
<math>\boxed{\textbf{(B)}}</math>.
 +
Note that when you write out a few numbers, you find that <math>f(n)=n+1</math> for any <math>n</math>, so <math>f(2017)=2018</math>
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2017|ab=A|num-b=6|num-a=8}}

Revision as of 00:10, 31 October 2021

Problem

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 $f(2017)$?

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

Solution

This is a recursive function, which means the function refers back to itself to calculate subsequent terms. 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 $\boxed{\textbf{(B)}}$. Note that when you write out a few numbers, you find that $f(n)=n+1$ for any $n$, so $f(2017)=2018$

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