2021 AIME II Problems/Problem 15: Difference between revisions
Etmetalakret (talk | contribs) No edit summary |
|||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
Let <math>f(n)</math> and <math>g(n)</math> be functions satisfying | |||
<cmath>f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ | |||
1 + f(n+1) & \text{ otherwise} | |||
\end{cases}</cmath>and | |||
<cmath>g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ | |||
2 + g(n+2) & \text{ otherwise} | |||
\end{cases}</cmath>for positive integers <math>n</math>. Find the least positive integer <math>n</math> such that <math>\tfrac{f(n)}{g(n)} = \tfrac{4}{7}</math>. | |||
==Solution== | ==Solution== | ||
We can't have a solution without a problem. | We can't have a solution without a problem. | ||
Revision as of 14:58, 22 March 2021
Problem
Let 
and 
be functions satisfying
![\[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\]](http://latex.artofproblemsolving.com/d/2/7/d27af3275e885b65d8cdb2caa1c2ef6bd46bd4a7.png)
and
![\[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\]](http://latex.artofproblemsolving.com/3/d/c/3dcced5eb55ddff1576102fc6df37624d1053bed.png)
for positive integers 
. Find the least positive integer such that 
.
Solution
We can't have a solution without a problem.
See also
| 2021 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America. 
