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, 2021

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:
Let $f(x) = x^{3} - 3x$ . Compute the number of positive divisors of
$\left\lfloor f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(\frac{5}{2}\right)\right)\right)\right)\right)\right)\right)\right)\right\rfloor$
where $f$ is applied 8 times.

Solution

Solution:
Note that $f\left(y + \frac{1}{y}\right) = \left(y + \frac{1}{y}\right)^{3} - 3\left(y + \frac{1}{y}\right) = y^{3} + \frac{1}{y^{3}}$ .

Thus, $f\left(2 + \frac{1}{2}\right) = 2^{3} + \frac{1}{2^{3}}$ , and in general $f^{k}\left(2 + \frac{1}{2}\right) = 2^{3^{k}} + \frac{1}{2^{3^{k}}}$ , where $f$ is applied $k$ times.

It follows that we just need to find the number of divisors of $\left\lfloor 2^{3^{8}} + \frac{1}{2^{3^{8}}} \right\rfloor = 2^{3^{8}}$ , which is just $3^{8} + 1 = 6562$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.