Difference between revisions of "2017 AMC 12A Problems/Problem 7"
(→Solution) |
(→Solution) |
||
Line 5: | Line 5: | ||
<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 <math>2</math> 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 <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> | ||
− | Alternatively, simply download java and run this method: | + | Alternatively, simply download java and run this method and input <math>n=2017</math>: |
public static int f(int n){ | public static int f(int n){ | ||
Line 18: | Line 18: | ||
return f(n-2) + 2; | return f(n-2) + 2; | ||
} | } | ||
+ | |||
+ | 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 12:49, 9 February 2017
Problem
Define a function on the positive integers recursively by ,
if
is even, and
if
is odd and greater than
. What is
?
Solution
This is a recursive function, which means the function is used to evaluate itself. To solve this, we must identify the base case, . We also know that when
is odd,
. Thus we know that
. Thus we know that n will always be odd in the recursion of
, and we add
each recursive cycle, which there are
of. Thus the answer is
. which is answer
Alternatively, simply download java and run this method and input :
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; }
It will return which is answer choice
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 |