Task B-1.1. Short acquaintances are not always the most reliable. That's why Barbara didn't immediately tell her new friend Luka her phone number directly, but instead satisfied his wish in a roundabout way: "I have two phones. The numbers of these phones are six-digit numbers with the following properties. In their decimal notation, the three-digit number at the beginning is half the three-digit number at the end. The sum of these two three-digit numbers is a number between 800 and 850. All three three-digit numbers are composed of 9 different digits."

Help Luka and determine which phone numbers are in question.
Solution. If the three-digit number at the beginning is $n$, then the three-digit number at the end is $2 n$, and their sum is $3 n$. There are a total of 17 multiples of three between 800 and 850, and if we exclude numbers with the same digits (822 and 828), there are 15. These numbers are

$$
801,804,807,810,813,816,819,825,831,834,837,840,843,846,849 \text {. }
$$

If we divide these numbers by three, we get the three-digit number $n$ that can be at the beginning, or the numbers

$$
267,268,269,270,271,272,273,275,277,278,279,280,281,282,283 .
$$

We will discard all numbers $n$ that do not have distinct digits, as well as those for which $n$ and $3 n$ do not have distinct digits. Only the numbers 267, 269, 273 remain. We will also discard the number 269 because then $2 n$, the three-digit number at the end, would be 538, which is impossible since it contains the digit 8, as does 3n. Therefore, the only six-digit numbers that satisfy all conditions are

$$
267534 \text { (sum 801) and } 273546 \text { (sum 819). }
$$