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

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==Problem==
 
==Problem==
  
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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>?
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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==
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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 <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
+
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
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<math>\boxed{\textbf{(B)}}</math>
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<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>
  
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Alternatively, simply download java and run this method and input <math>n=2017</math>:
+
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/ZxcTc-3FoiU
  
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    public static int f(int n){
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~Education, the Study of Everything
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        if(n == 1)
 
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            return 2;
 
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        else if(n % 2 == 0)
 
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            return f(n-1) + 2;
 
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        else
 
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            return f(n-2) + 2;
 
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    }
 
−
 
 
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It will return <math>f(2017) = 2018</math> which is answer choice <math>\boxed{\textbf{(B)}}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=6|num-a=8}}
 
{{AMC12 box|year=2017|ab=A|num-b=6|num-a=8}}

Revision as of 13:59, 10 June 2023

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 $f(2017)=2018$

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/ZxcTc-3FoiU

~Education, the Study of Everything

See Also

2017 AMC 12A (Problems • Answer Key • Resources)
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
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