Fomin S.V.

Among ten-digit numbers, which are there more of: those that can be represented as the product of two five-digit numbers, or those that cannot be represented this way?
The number of ten-digit numbers representable as the product of two five-digit numbers is no more than the number of all pairs of five-digit numbers (including pairs of identical numbers). (In fact, there are significantly fewer: firstly, not all products of two five-digit numbers will be ten-digit, and secondly, different pairs of numbers can yield the same product.) Therefore, it is sufficient to check that the number of pairs of five-digit numbers is less than half the number of ten-digit numbers.

The first method. There are $9 \cdot 10^{9}$ ten-digit numbers in total (see the solution to problem 60336), and twice the number of pairs of five-digit numbers is

equal to $9 \cdot 10^{4}\left(9 \cdot 10^{4}+1\right)=81 \cdot 10^{8}+9 \cdot 10^{4}<9 \cdot 10^{9}$.

The second method. Each pair $(A, B)$ of two different five-digit numbers can be associated with two ten-digit numbers $\overline{A B}$ and $\overline{B A}$ (by writing them one after the other). Each pair of the form $(A, B)$ can also be associated with two numbers: $\overline{A A}$ and $\overline{A 00000}$.

Note that this does not cover all ten-digit numbers: the number 1111101111 cannot be obtained this way. Therefore, the number of pairs of five-digit numbers is less than half the number of ten-digit numbers.

## Answer

There are more non-representable.