We can solve this problem using the formula for the future value of a series of regular deposits into a savings account with compound interest:

$A = P \times \frac{(1 + r/n)^{nt} - 1}{r/n}$

Where:

• $A$ is the future value (in this case, $100,000),
• $P$ is the monthly deposit (in this case, $200),
• $r$ is the annual interest rate (in this case, 2.1%, or 0.021),
• $n$ is the number of times the interest is compounded per year (since this is monthly, $n = 12$),
• $t$ is the time in years, which we are solving for.

Let’s rearrange the formula to solve for $t$:

$t = \frac{\log \left( \frac{A \cdot r/n + P}{P} \right)}{n \cdot \log(1 + r/n)}$

Now, let’s substitute the known values into the equation:

• $A = 100,000$,
• $P = 200$,
• $r = 0.021$,
• $n = 12$.

I'll calculate the time $t$ for you.It will take approximately 29.96 years, or nearly 30 years, for your savings account to reach $100,000 with a monthly deposit of $200 at an annual interest rate of 2.1% compounded monthly.

Would you like further details, or do you have any other questions?

Here are five related questions:

1. What if you increase the monthly deposit to $300? How long would it take?
2. How does the result change if the interest rate is 3% instead of 2.1%?
3. What if you want to accumulate $200,000 instead of $100,000? How long would that take?
4. What is the total amount of interest earned over the 30 years?
5. How does the frequency of compounding (monthly vs yearly) affect the time to reach $100,000?

Tip: The more frequent the compounding, the faster your savings grow due to the power of compound interest.