3. How many five-digit numbers exist where the sum of the first two digits is twice the sum of the last two digits?
Answer: 2600.

Solution. Let the first two digits of the number be $a$ and $b$, and the last two digits be $c$ and $d$. We need to find the number of combinations of these digits for which $a+b=2(c+d)$. Since $a+b \leqslant 18$, then $c+d \leqslant 9$.

Let $x_{i}(i=1,2, \ldots, 9)$ denote the number of pairs of the last two digits whose sum is $i$. The first digit can take values from 0 to $i$, after which the second digit is uniquely determined. Thus, $x_{i}=i+1$.

Let $y_{i}(i=1,2, \ldots, 9)$ denote the number of pairs of the first two digits whose sum is $2i$. Each time, the digit $b$ is uniquely determined by the digit $a$: $b=2i-a$. Therefore, we need to count the number of possible values for the digit $a$. If $i \leqslant 4$, i.e., $2i \leqslant 8$, then the digit $a$ can take values from 1 to $2i$ - a total of $2i$ values. In other words, in this case, $y_{i}=2i$. If $5 \leqslant i \leqslant 9$, i.e., $10 \leqslant 2i \leqslant 18$, then the digit $a$ can take values from $2i-9$ to $9$ - a total of $19-2i$ values, and $y_{i}=19-2i$.

Let's compile the found values of $x_{i}$ and $y_{i}$ into one table.

| $i$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x_{i}$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| $y_{i}$ | 2 | 4 | 6 | 8 | 9 | 7 | 5 | 3 | 1 |

The number of five-digit numbers in which $c+d=i$ and $a+b=2i$ is $10 x_{i} y_{i}$ (the middle digit can be chosen in 10 ways). Therefore, the total number of the desired numbers is $10\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{9} y_{9}\right)$. Substituting the found values gives the answer 2600.

Evaluation. 13 points for a complete solution. If all $x_{i}$ and $y_{i}$ are correctly found but there is an arithmetic error in the final calculation, 11 points. If the solution approach is correct but some $x_{i}$ or $y_{i}$ are incorrectly determined, 4-5 points.