Slovenija 2008

Let us compare the expressions $\frac{{\log }_{10}2{\log }_{10}3\cdots {\log }_{10}(n-1)}{10^{n-2}}$ and $\frac{{\log }_{10}2{\log }_{10}3\cdots {\log }_{10}n}{10^{n-1}}$. The inequality $$\frac{{\log }_{10}2{\log }_{10}3\cdots {\log }_{10}(n-1)}{10^{n-2}}\ge \frac{{\log }_{10}2{\log }_{10}3\cdots {\log }_{10}n}{10^{n-1}}$$ holds if and only if $1\ge \frac{1}{10}{\log }_{10}n={\log }_{10}\sqrt[10]{n}$, which is equivalent to $10\ge \sqrt[10]{n}$ and $10^{10}\ge n$. This implies $$\begin{array}{rl}\frac{{\log }_{10}2}{10}& >\frac{{\log }_{10}2\cdot {\log }_{10}3}{10^2}>\cdots >\frac{{\log }_{10}2\cdot {\log }_{10}3\cdots {\log }_{10}(10^{10}-1)}{10^{10^{10}-2}}\\ & =\frac{{\log }_{10}2\cdot {\log }_{10}3\cdots {\log }_{10}(10^{10})}{10^{10^{10}-1}}\end{array}$$ and $$\begin{array}{rl}\frac{{\log }_{10}2\cdot {\log }_{10}3\cdots {\log }_{10}(10^{10})}{10^{10^{10}-1}}<& \frac{{\log }_{10}2\cdot {\log }_{10}3\cdots {\log }_{10}(10^{10}+1)}{10^{10^{10}}}\\ <& \frac{{\log }_{10}2\cdot {\log }_{10}3\cdots {\log }_{10}(10^{10}+2)}{10^{10^{10}+1}}<\dots \end{array}$$ So, the expression $\frac{{\log }_{10}2{\log }_{10}3\cdots {\log }_{10}n}{10^{n-1}}$ has the smallest possible value when $n=10^{10}-1$ or $n=10^{10}$. For these two values of $n$ the value of the expression is $$\frac{{\log }_{10}2{\log }_{10}3\cdots {\log }_{10}10^{10}}{10^{10^{10}-1}}$$