Difference between revisions of "2017 AMC 12A Problems/Problem 7"
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==Problem== | ==Problem== | ||
| − | Define a function on the positive integers recursively by <math>f(1) = 2</math>, <math>f(n) = f(n-1) + | + | 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> | ||
Revision as of 09:02, 16 April 2019
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
.
Note that when you write out a few numbers, you find that 
for any , so 
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 | |
