1. Which whole numbers from 1 to $4 \cdot 10^{25}$ (inclusive) are there more of, and by how many: those containing only even digits or those containing only odd digits?
Answer: the number of numbers containing only odd digits is greater by $\frac{5^{26}-5}{4}$.

Solution. Consider $k$-digit numbers ( $1 \leqslant k \leqslant 25$ ). The number of numbers consisting only of odd digits is $5^{k}$ (for each of the $k$ positions, any of the digits $1,3,5,7$, 9 can be chosen); the number of numbers consisting only of even digits is $4 \cdot 5^{k-1}$ (for the first position, any of the digits $2,4,6,8$ can be chosen, and for the remaining positions - any of the digits $0,2,4,6$, 8). Therefore, the number of $k$-digit numbers containing only odd digits is $5^{k}-4 \cdot 5^{k-1}=5^{k-1}$ more than the number of numbers containing only even digits.

Consider 26-digit numbers not exceeding $4 \cdot 10^{25}$. Among them, the number of numbers written only with odd digits is $2 \cdot 5^{25}$, and the number of numbers written only with even digits is $-.5^{25}+1$ (do not forget to account for the number $\left.4 \cdot 10^{25}\right)$.

Thus, the number of numbers written only with odd digits is greater by $1+5+5^{2}+5^{3}+\ldots+5^{24}+5^{25}-1=$ $\frac{5^{26}-1}{5-1}-1=\frac{5^{26}-5}{4}$.