Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

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.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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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