Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Prove it Philippines

Problem:

Let $f(m) = 2^{2^{2 \cdots}}$ ( $m$ times). Find the least $m$ so that $\log_{10} f(m)$ exceeds $6$ .

Solution

Solution:

$f(4) = 2^{16} = 65536 < 10^{6} < 2^{65536} = f(5)$ .

Therefore, the least $m$ is $5$ .

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