Problem 4.7. On the table, there are 4 stacks of coins. The first stack has 9 coins, the second has 7, the third has 5, and the fourth has 10. In one move, it is allowed to add one coin to three different stacks. What is the minimum number of moves required to make the number of coins in all stacks equal?
Answer: 11.

Solution. Suppose $N$ moves were made, after which the number of coins in all stacks became equal.

Let's slightly change the rules. Suppose initially there were not 9, 7, 5, and 10 coins in the stacks, but $N+9, N+7, N+5$, and $N+10$ respectively; and we will perform the moves as follows: instead of adding one coin to three stacks, we will take one coin from a stack (the one into which we did not add a coin during the original move). Note that the final result will not change! (In fact, instead of adding coins to three stacks, we add them to all four and then take one away.)

Under the new rules, answering the question is much easier. In one move, we take one coin from any stack, and our goal is to make the number of coins in all stacks equal as quickly as possible. It is easy to understand that for this, we need to leave $N+5$ coins in each stack. For this, we need to take 4 coins from the first stack, 2 from the second, 0 from the third, and 5 from the fourth. In total, we need to make $4+2+0+5=11$ moves.