25.24. Let among the numbers $2^{1}, 2^{2}, \ldots, 2^{n}$ exactly $a_{n}$ numbers start with the digit 1. Compute the limit $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}$.
25.24. Consider all natural numbers $a$ for which $10^{k-1} \leqslant a10^{k}$. The smallest power of two within these limits starts with a one, since otherwise we could divide the number $a=2^{m}$ by 2 and obtain a number within the same limits. The power of two following the smallest one starts with the digit 2 or 3. Therefore, among the considered numbers, there is exactly one power of two starting with one. Thus, if $10^{k-1} \leqslant 2^{n}<10^{k}$, then $a_{n}=k-1$. Therefore, $a_{n} \leqslant n \lg 2 < a_{n}+1$, i.e., $\lg 2 - \frac{1}{n} < \frac{a_{n}}{n} \leqslant \lg 2$. Hence, $\lim _{n \rightarrow \infty} \frac{a_{n}}{n} = \lg 2$.