1. Which whole numbers from 1 to 60000 (inclusive) are there more of, and by how many: those containing only even digits or those containing only odd digits?
Answer: There are 780 more numbers containing only odd digits.

Solution. Consider $k$-digit numbers ( $1 \leqslant k \leqslant 4$ ). 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, there are $5^{k}-4 \cdot 5^{k-1}=5^{k-1}$ more $k$-digit numbers containing only odd digits in their notation than numbers containing only even digits.

Consider 5-digit numbers not exceeding 60000. Among them, the number of numbers written only with odd digits is $3 \cdot 5^{4}$, and the number of numbers written only with even digits is $2 \cdot 5^{4}+1$ (do not forget to account for the number 60000 itself).

Thus, there are $1+5+5^{2}+5^{3}+5^{4}-1=780$ more numbers written only with odd digits.