Interchangeable percentages

Rewriting this equation as fractions:

$(10^n)\% \div (n^{10})\% = 10 \Rightarrow \frac{10^n}{100}\div \frac{n^{10}}{100}=10$

$\frac{10^n}{100}\div \frac{n^{10}}{100}=\frac{10^n}{100}\cdot \frac{100}{n^{10}}=10$

$\frac{10^n\cdot 10^2}{n^{10}\cdot 10^2}=10$

$\frac{10^n}{n^{10}}=10$

$10^n=10\cdot n^{10}$

$10^{n-1}=n^{10}$

$\log{n^{10}}=n-1$

$10\cdot\log{n}=n-1$

$\log{n}=\frac{n-1}{10}$

We are looking for a positive integer, which one is $n$ . If $n$ is a integer, then $n-1$ is a integer, which means that $\log{n}$ is rational because $\frac {n-1}{10}$ is rational too. With these restrictions, the only possible numbers for $n$ is powers of 10. But 10, 100, 1000 or higher, aren't solutions for this equation. Then, 1 is the only number that satisfies these conditions, and $\log{1}=0$ , which works when $\frac{n-1}{10}=0$ .

Then the answer is $\boxed{1}$ .