The problem requires us to find the percentage of girls who did not pass the examination in a class where 60% are girls, and the difference in number between girls and boys and passing information is provided.
\( 0.2N = 30 \). Thus, \( N = \frac{30}{0.2} = 150 \).
\( \frac{68}{100} \times 150 = 102 \).
\( 102 - 30 = 72 \).
\( \frac{18}{90} \times 100 = 20\% \).
Thus, the percentage of the girls who do not pass is 20%.
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: