The Problem was as follows:
Define $\log*(n)$ to be the smallest number of times the log function must be iteratively applied to $n$ to get a result less than or equal to $1$. For example $\log*(1000)=2$ since $log(1000)=3$ and $log(3)=0.477 \le 1 $. Let $a$ be the smallest integer such that $\log*(a)=3$. Computer the number of zeroes in the base 10 representation of $a$
My answer was $10^{10}$ but they claimed it to be $9$
My logic was that $\log10^{10^{10}}=10^{10}$ and $\log10^{{10}}={10}$ and $\log10=1$ and $1\le1$ so the $\log*{10^{10^{10}}}=3$.
I presume this means that a smaller answer exists, or I made a logic error somewhere. Can someone show how the smallest answer has 9 zeroes?
$10^{10^{10}}$ is not the smallest correct response. Notice that $10^{10}+1$ also succeeds: $\log \log{10^{10}} = \log 10 = 1$, so $\log \log(10^{10} + 1) > 1$. Hence 3 logs are needed to reduce $10000000001$ to a number less than or equal to one, since two is not enough. A simple count gives 9 zeroes.