Question: A 25-year-old worker plans to retire at age 65. He believes that $800,000 is needed to retire comfortably. How much should be deposited now at 4% compounded monthly to meet the $800,000 retirement goal?

1. **State the problem:** We want to find the present deposit amount $P$ that will grow to $800,000$ in $40$ years (from age $25$ to $65$) with an interest rate of $4\%$ compounded monthly. 2. **Formula used:** The formula for compound interest is: $$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$ where: - $A$ is the amount of money accumulated after $t$ years, including interest. - $P$ is the principal (the initial deposit). - $r$ is the annual interest rate (decimal). - $n$ is the number of times interest is compounded per year. - $t$ is the time the money is invested for in years. 3. **Given values:** - $A = 800000$ - $r = 0.04$ - $n = 12$ (monthly compounding) - $t = 40$ 4. **Rearrange the formula to solve for $P$:** $$ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} $$ 5. **Calculate the denominator:** $$ 1 + \frac{0.04}{12} = 1 + 0.0033333333 = 1.0033333333 $$ 6. **Calculate the exponent:** $$ nt = 12 \times 40 = 480 $$ 7. **Calculate the compound factor:** $$ \left(1.0033333333\right)^{480} $$ 8. **Calculate the compound factor value:** $$ \left(1.0033333333\right)^{480} \approx 4.80102 $$ 9. **Calculate the present deposit $P$:** $$ P = \frac{800000}{4.80102} \approx 166,621.68 $$ 10. **Round up to the nearest dollar:** $$ P = 166,622 $$ **Final answer:** The worker should deposit **$166,622$** now to reach $800,000$ in 40 years at 4% interest compounded monthly.