Difference between revisions of "2021 AIME II Problems/Problem 15"

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==Problem==
 
==Problem==
These problems will not be posted until the 2021 AIME II is released on Thursday, March 25, 2021.
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Let <math>f(n)</math> and <math>g(n)</math> be functions satisfying
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<cmath>f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\
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1 + f(n+1) & \text{ otherwise}
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\end{cases}</cmath>and
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<cmath>g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\
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2 + g(n+2) & \text{ otherwise}
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\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>.
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==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 $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\]and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\]for positive integers $n$. Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$.

Solution

We can't have a solution without a problem.

See also

2021 AIME II (ProblemsAnswer KeyResources)
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Problem 14
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