We can use the formula for compound interest to solve this problem. The formula is:

$A = P \left( 1 + \frac{r}{n} \right)^{nt}$

Where:

• $A$ is the amount of money accumulated after $t$ years (future value),
• $P$ is the principal (the initial amount of money to be deposited),
• $r$ is the annual interest rate (in decimal form),
• $n$ is the number of times the interest is compounded per year,
• $t$ is the time the money is invested or borrowed for, in years.

Given:

• $A = 18,822$ (the future value),
• $r = 4.5\%$ or $0.045$ (the annual interest rate),
• $n = 12$ (since interest is compounded monthly),
• $t = 11$ years.

We need to solve for $P$ (the principal). Rearranging the compound interest formula to solve for $P$:

$P = \frac{A}{\left( 1 + \frac{r}{n} \right)^{nt}}$

Now substitute the given values:

$P = \frac{18,822}{\left( 1 + \frac{0.045}{12} \right)^{12 \times 11}}$

Let's calculate this step-by-step:

$P = \frac{18,822}{\left( 1 + 0.00375 \right)^{132}} = \frac{18,822}{(1.00375)^{132}}$

First, calculate $(1.00375)^{132}$, then divide 18,822 by that result.

$P \approx \frac{18,822}{1.62889} \approx 11,554.50$

So, you would need to deposit approximately $11,554.50 now in order to have $18,822 in 11 years.

Would you like more details on any step or have any questions?

Here are some related questions:

1. How would the required deposit change if the interest rate were higher?
2. What would happen if the compounding period was quarterly instead of monthly?
3. How much would you have if you deposited $12,000 now at the same interest rate for 11 years?
4. How can we calculate the interest earned after 5 years under these conditions?
5. What would be the future value after 11 years if no additional deposits were made?

Tip: The more frequently interest is compounded, the more total interest you'll earn, increasing the final amount.