3. 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, you are 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?

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# Answer: in 11 moves.

Solution 1. Consider the differences between the number of coins in each of the other stacks and the number of coins in stack No. 3. In one move, either each of these differences increases by 1 (if all stacks except 3 are chosen for adding coins), or exactly 1 of these differences decreases by 1 (in other cases). Initially, these differences are 4, 2, and 5. Thus, to reduce all these differences to 0, it will take no less than $4+2+5=11$ moves.

11 moves are possible: 4 times we choose all stacks except No. 1, 2 times all stacks except No. 2, and 5 times all stacks except No. 4. In the end, there will be 16 coins in each stack.

Solution 2. Let us say we have made $s$ moves, of which $s_{\mathrm{k}}$ moves involve adding a coin to all stacks except $\mathrm{k}(\mathrm{k} \in\{1,2,3,4\})$. Then, from the condition that the number of coins in all stacks is equal after all moves, we get: $9+\mathrm{s}-\mathrm{s}_{1}=7+\mathrm{s}-\mathrm{s}_{2}=5+\mathrm{s}-\mathrm{s}_{3}=10+\mathrm{s}-\mathrm{s}_{4}$. Therefore, $s_{1}=4+s_{3} \geq 4, s_{2}=2+s_{3} \geq 2$ and $s_{4}=5+s_{3} \geq 5$. Thus, $s \geq s_{1}+s_{2}+s_{4} \geq 4+2+5=11$. An example of 11 moves is the same as in Solution 1.

Criteria. Only the answer - 1 point. Answer with an example - 3 points. Only proof that the number of moves is no less than 11 - 4 points.