Problem 7.3. The miserly knight has 5 chests of gold: the first chest contains 1001 gold coins, the second - 2002, the third - 3003, the fourth - 4004, the fifth - 5005. Every day, the miserly knight chooses 4 chests, takes 1 coin from each, and places them in the remaining chest. After some time, there were no coins left in the first chest, and there were exactly 2019 coins in another chest. Which one?
Answer: in the fifth.

Solution. Let's look at the remainders of the number of coins in the chests when divided by 5. Each day, the number of coins in each chest either decreases by 1 or increases by 4, which means the remainder when divided by 5 always decreases by 1 (if it was 0, it will become 4). Since after some time there were no coins left in the first chest, the number of days that have passed is $5k+1$ for some natural number $k$, and in each chest, the remainder has decreased by 1. Since 2019 gives a remainder of 4 when divided by 5, such a number of coins could only remain in the fifth chest.

## Criteria

The largest suitable criterion is used:

76. Any correct solution to the problem.

## 3 6. The idea of considering remainders when divided by 5 is present.

## 16. The correct answer is given.