Fomin S.B.

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

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The number of ten-digit numbers that can be represented 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 numbers, 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 a total of $9 \cdot 10^{9}$ ten-digit numbers (see the solution to problem $\underline{60336}$), and twice the number of pairs of five-digit numbers

is $9 \cdot 10^{4}\left(9 \cdot 10^{4}+1\right)=81 \cdot 10^{8}+9 \cdot 10^{4}$. Problem $\underline{98025}$ Topics:

$\left[\begin{array}{ll}\text { equations in integers } \\ \text { equations in integers }\end{array}\right]$

## Author: Folklore

Find the number of solutions in natural numbers to the equation $\left[{ }^{x} / 10\right]=[x / 11]+1$.

## Solution

Let $x=11 n+r$, where $n \geq 0,0 \leq r \leq 10$. Then $\left[{ }^{x} / 11\right]=n, n+1=\left[{ }^{x} / 10\right]=n+\left[{ }^{n+r} / 10\right]$, which means $10 \leq n+r<20,10$ $-r \leq n \leq 19-r$. For each $r$ from 0 to 10, we get 10 solutions.

## Answer

110 solutions.