3. (3 points) Anya, Vanya, Danya, and Tanya were collecting apples. It turned out that each of them collected a whole percentage of the total number of apples collected, and all these numbers were different and greater than zero. Then Tanya, who collected the most apples, ate her apples. After this, it turned out that each of the children still had a whole percentage, but now of the remaining number of apples. What is the minimum number of apples that could have been collected?

Answer: 20 : for example $2+3+5+10$
# Solution:

Let Anya have collected the fewest apples, which is $k$, and the total number of apples is $n$. Then we get that the total number of apples collected is no less than $4k + 6$. For $k \geq 5$, we already get more than 20 apples. For $k < 5$, we need to consider the cases where $k$ is 1, 2, 3, or 4. 

If $k = 4$, then the total number of apples is at least $4 \times 4 + 6 = 22$, which is more than 20. 

If $k = 3$, then the total number of apples is at least $4 \times 3 + 6 = 18$, which is less than 20. However, since $\frac{n}{k}$ must be an integer and a divisor of 100, we get that at least $10k$ apples were collected, which is 30, more than 20.

If $k = 2$, then the total number of apples is at least $4 \times 2 + 6 = 14$, which is less than 20. However, since $\frac{n}{k}$ must be an integer and a divisor of 100, we get that at least $10k$ apples were collected, which is 20.

If $k = 1$, we get that Anya, Vanya, and Sanya collected at least 6 apples, and thus, by similar reasoning, there are at least 10 of them. Therefore, the four of them together had at least 20 apples (the next divisor of 100).