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==
-Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?
-<math>\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%</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 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>.
+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>
+==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
-Let <math>j</math> represent how far Jerry walked, and <math>s</math> represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides,
+==See Also==
-<math>j = 2</math>
+{{AMC12 box|year=2017|ab=A|num-b=6|num-a=8}}
-Since Silvia walked the diagonal, she walked the hypotenuse of a 45, 45, 90 triangle with leg length 1. Thus,
-<math>s = \sqrt{2} = 1.414...</math>
-We can then take
-<math>\frac{j-s}{j} = \frac{\sqrt{2 - 1.4}}{2} = 0.3 = 30\%</math>
-<math>\boxed{ \textbf{A}}</math>.

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