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

Latest revision as of 06:23, 14 September 2024

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

Video Solution

https://www.youtube.com/watch?v=Vaauk0gNy_k

~Math4All999

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

@@ Line 1: / Line 1: @@
 ==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>
 ==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>
-Alternatively, simply download java and run this method:
+==Video Solution (HOW TO THINK CREATIVELY!!!)==
+https://youtu.be/ZxcTc-3FoiU
+~Education, the Study of Everything
-    public static int f(int n){
+==Video Solution==
-        if(n == 1)
+https://www.youtube.com/watch?v=Vaauk0gNy_k
-            return 2;
-        else if(n % 2 == 0)
+~Math4All999
-            return f(n-1) + 2;
-        else
-            return f(n-2) + 2;
-    }
 ==See Also==
 {{AMC12 box|year=2017|ab=A|num-b=6|num-a=8}}

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