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

(See Also)
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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 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 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>.
 
<math>\boxed{\textbf{(B)}}</math>
 
<math>\boxed{\textbf{(B)}}</math>
 +
 +
Alternatively, simply download java and run this program:
 +
public class Amc12a7
 +
{
 +
    public static int f(int n){
 +
        if(n == 1)
 +
            return 2;
 +
        else if(n % 2 == 0)
 +
            return f(n-1) + 2;
 +
        else
 +
            return f(n-2) + 2;
 +
    }
 +
}
  
 
==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}}
{{MAA Notice}}
 

Revision as of 11:34, 9 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 $1008*2+2=2018$. $\boxed{\textbf{(B)}}$

Alternatively, simply download java and run this program: public class Amc12a7 {

   public static int f(int n){
       if(n == 1)
           return 2;
       else if(n % 2 == 0)
           return f(n-1) + 2;
       else
           return f(n-2) + 2;
   }

}

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