Problem 3. Find a six-digit number where the first digit is 6 times less than the sum of all digits to the right of it and the second digit is 6 times less than the sum of all digits to the right of it.

$[4$ points] (A. V. Shapovalov)
Answer: 769999.

First solution. The sum of the last five digits of the number, by condition, is divisible by 6 (the quotient is equal to the first digit). The sum of the last four digits of the number is also divisible by 6 (the quotient is equal to the second digit). Therefore, the second digit, as the difference of two numbers divisible by 6, is also divisible by 6. This means the second digit is 0 or 6. If it is 0, then the sum of the last four digits is also 0, and thus the first digit is 0, but a six-digit number cannot start with 0. Therefore, this option does not work, and the second digit is 6. Then the sum of the last four digits is $6 \cdot 6 = 36$, so all of them are 9. Finally, the sum of the last five digits is 42, and thus the first digit is $42 : 6 = 7$. Therefore, the required number is 769999.

Second solution. The sum of the last 4 digits is 6 times the second digit, so the sum of the last 5 digits is 7 times the second digit and 6 times the first digit. Therefore, 7 times the second digit equals 6 times the first digit, which means the first digit is 7, and the second digit is 6; then the sum of the last four is 36, and this is only possible if all 4 last digits are nines.

Comment. The problem can be solved by trying all possible values (from 0 to 9) of the second digit of the number. For example, if this digit is 9, then the sum of the digits to the right of it is 6 times greater, i.e., $9 \cdot 6 = 54$, but then the sum of all digits to the right of the first is $9 + 54 = 63$. Since the number 63 is not divisible by 6, the option that the second digit is 9 does not work, and so on.