Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Find the answer

Find the smallest positive integer $n$ such that $\underbrace{2^{2 \cdot 2}}_{n}>3^{3^{3^{3}}}$ . (The notation $\underbrace{2^{2^{2}}}_{n}$ , is used to denote a power tower with $n 2$ 's. For example, $\underbrace{2^{22^{2}}}_{n}$ with $n=4$ would equal $2^{2^{2^{2}}}$ .)

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Clearly, $n \geq 5$ . When we take $n=5$ , we have $2^{2^{2^{2^{2}}}}=2^{2^{16}}<3^{3^{27}}=3^{3^{3^{3}}}.$ On the other hand, when $n=6$ , we have $2^{2^{2^{2^{2^{2}}}}}=2^{2^{65536}}=4^{2^{65535}}>4^{4^{27}}>3^{3^{27}}=3^{3^{3^{3}}}.$ Our answer is thus $n=6$ .

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