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Which five-digit numbers are more numerous: those not divisible by 5 or those in which neither the first nor the second digit from the left is a five?
Let's calculate the number of numbers not divisible by 5. The first digit can be any of the 9 non-zero digits, the second, third, and fourth digits can be any of the 10 digits, and the last digit can be any of the eight digits that are not 5 or 0. In total, we get $9 \cdot 8 \cdot 10^{3}$ numbers (cf. problem $\underline{60366}$).

Now let's calculate the number of numbers where neither the first nor the second digit is a 5. The first digit can be any of the 8 digits (not 5 and not 0), the second digit can be any of the nine digits (not 5), and the third, fourth, and fifth digits can be any of the 10 digits. We get the same result: $9 \cdot 8 \cdot 10^{3}$ numbers.

## Answer

Equally.