Let the initial amount of milk in the container be $M_0 = 1$ (since we are dealing with proportions, we can assume the total amount of liquid in the container is 1 unit).

First replacement:

• After the first replacement, 10% of the milk is removed, which means $90\%$ of the milk remains.
• Therefore, after the first step, the amount of milk in the container is: $M_1 = 0.9 \times M_0 = 0.9$

Second replacement:

• After the second replacement, 10% of the solution (which now consists of milk and water) is removed. Since $90\%$ of the current solution remains, the amount of milk left is $90\%$ of the milk from the previous step: $M_2 = 0.9 \times M_1 = 0.9 \times 0.9 = 0.9^2$

General case:

• After $n$ replacements, the amount of milk left in the container is: $M_n = 0.9^n$

The amount of water in the container after $n$ steps is $1 - M_n$.

Condition for more water than milk:

We are given that after $n$ replacements, there is more water than milk. Thus, the condition is: $1 - M_n > M_n$ Substituting $M_n = 0.9^n$ into this inequality: $1 - 0.9^n > 0.9^n$ Simplifying: $1 > 2 \times 0.9^n$ $0.5 > 0.9^n$ Taking the logarithm on both sides: $\log(0.5) > n \log(0.9)$ $n > \frac{\log(0.5)}{\log(0.9)}$ Using approximations for the logarithms: $\log(0.5) \approx -0.3010 \quad \text{and} \quad \log(0.9) \approx -0.0458$ Thus: $n > \frac{-0.3010}{-0.0458} \approx 6.57$ Since $n$ must be an integer, the minimum value of $n$ is: $n = 7$

Thus, the minimum value of $n$ is $\boxed{7}$.

Would you like more details on the solution, or have any further questions?

Here are 5 related questions you might explore:

1. What would happen if only 5% of the solution was replaced each time?
2. How would the formula change if the percentage of milk removed varied in each step?
3. Can this process ever result in all water and no milk after a finite number of steps?
4. How does the rate of decrease of milk change with each successive step?
5. What is the total percentage of milk remaining after 3 replacements?

Tip: Understanding how exponential decay works can help solve problems where quantities decrease proportionally over repeated steps.