Difference between revisions of "2017 AMC 12A Problems/Problem 7"
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<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> | 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> | ||
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+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/ZxcTc-3FoiU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==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}} |
Latest revision as of 14:59, 10 June 2023
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 refers back to itself to calculate subsequent terms. 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
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
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 |