2. A six-digit number $A$ is divisible by 19. The number obtained by removing its last digit is divisible by 17, and the number obtained by removing the last two digits of $A$ is divisible by 13. Find the largest $A$ that satisfies these conditions.
2. Answer: 998412.

Solution. The largest four-digit number divisible by 13 is 9997. Among the numbers from 99970 to 99979, there is a number 99977 divisible by 17, but among the numbers from 999770 to 999779, there is no number divisible by 19. However, for the next number 9984, which is divisible by 13, the number 99841 is divisible by 17, and the number 998412 is divisible by 19.

Criteria. Correct and justified answer - 7 points; correct answer, but not proven to be the largest - 4 points; incorrect answer - 0 points.