Tens Everywhere

Let $y= \log _{10} \left(\dfrac{x}{10} \right)$ . Note that $\log _{ \left(\frac{x}{10} \right) } 10 = \frac{1}{y}$ . Let's rewrite the given inequality using these substitutions.

$\begin{aligned} \frac{1} {10} \frac{1}{y} & > 10 y \\ 0 & > 10y - \frac{1}{10 y } \\ 10y - \frac{1}{10 y } & < 0 \\ \frac{(10y)^{2} - 1 }{10y} & < 0 \\ \frac{(10y-1)(10y+1)}{10y} & < 0 \end{aligned}$

This means that either $y < -\frac{1}{10}$ or $0 < y < \frac{1}{10}$ .

$\begin{array}{rl|l} y &< -\frac{1}{10} & 0 < y < \frac{1}{10} \\ \log _{10} \left(\frac{x}{10} \right) &< - \frac{1}{10} & 0 < \log _{10} \left(\frac{x}{10} \right) < \frac{1}{10} \\ \frac{x}{10} &< 10 ^{-\frac{1}{10}} & 10^{0} < \frac{x}{10} < 10^{\frac{1}{10}} \\ x &< 10^{1 - \frac{1}{10} } & 1 < \frac{x}{10} < 10^{\frac{1}{10}} \\ x &< 10^{\frac{9}{10}} & 10 < x < 10^{\frac{11}{10}} \\ x &< 7.94 \ldots & 10 < x < 12.589 \ldots \\ \end{array}$

The positive integer solutions satisfying the first inequality are $\{ 1, 2, 3, 4, 5, 6, 7\}$ and the positive integers satisfying the second inequality are $\{ 11, 12 \}$ .

If either of the inequalities is satisfied, then the original inequality in the problem becomes true, therefore we will take the union of both the inequalities. The integer solutions are $\{1, 2, 3, 4, 5, 6, 7, 11, 12 \}$ . The sum of all the solutions is $\boxed{51}$ $_\square$

To simplify the notation, let $X = \frac{x}{10}$ . So X is a positive integer multiple of 0.1. The equation becomes $\frac{\log_X 10}{10} > 10 \log_{10} X \\ \Rightarrow \frac{\lg 10}{10 \lg X} > \frac{10 \lg X}{\lg 10} \\ \Rightarrow 100 \lg X < \frac{1}{\lg X}$

Obviously $\lg X \neq 0$ . Thus there are two cases to consider.

Case 1: $\lg X < 0$ $100 (\lg X)^2 > 1 \\ \Rightarrow (\lg X)^2 > 0.01 \\ \Rightarrow -\lg X > 0.1 \\ \Rightarrow \lg X < -0.1 \\ \Rightarrow X < 10^{-0.1} \\ \Rightarrow x < 10^{0.9} \\ \Rightarrow x \in \{1, 2, 3, 4, 5, 6, 7\}$ (The fourth step naturally verifies that the case premise is satisfied.)

Case 2: $\lg X > 0$ $100 (\lg X)^2 < 1 \\ \Rightarrow (\lg X)^2 < 0.01 \\ \Rightarrow \lg X < 0.1$ Combining this with the case premise: $0 < \lg X < 0.1 \\ \Rightarrow 1 < X < 10^{0.1} \\ \Rightarrow 10 < x < 10^{1.1} \\ \Rightarrow x \in \{11, 12\}$

Combining the two cases, $x \in \{1, 2, 3, 4, 5, 6, 7, 11, 12\}$ .