1. **Problem Statement:** We want to find how long it takes to save at least 13000 by making deposits of 900 at the end of every 6 months in an account earning 3.66% interest compounded quarterly. 2. **Identify the formula:** This is a future value of an annuity problem with compound interest. The formula for the future value $FV$ of an annuity with periodic deposits $P$, interest rate per period $i$, and number of periods $n$ is: $$FV = P \times \frac{(1+i)^n - 1}{i}$$ 3. **Important details:** - Deposits are every 6 months (semiannual). - Interest is compounded quarterly (every 3 months). - Annual nominal interest rate is 3.66%. 4. **Adjust interest rate and periods:** - Quarterly interest rate $i_q = \frac{3.66}{100} \div 4 = 0.00915$ per quarter. - Deposits every 6 months means every 2 quarters. 5. **Effective interest rate per deposit period:** Since deposits are every 2 quarters, the effective interest rate per deposit period is compounded twice quarterly: $$i = (1 + i_q)^2 - 1 = (1 + 0.00915)^2 - 1 = 0.01837$$ 6. **Set up the equation:** We want $FV \geq 13000$, with $P=900$, $i=0.01837$, and $n$ unknown (number of 6-month periods): $$13000 \leq 900 \times \frac{(1+0.01837)^n - 1}{0.01837}$$ 7. **Solve for $n$:** $$\frac{13000 \times 0.01837}{900} + 1 \leq (1.01837)^n$$ Calculate left side: $$\frac{13000 \times 0.01837}{900} + 1 = \frac{238.81}{900} + 1 = 0.26534 + 1 = 1.26534$$ Take natural logarithm: $$\ln(1.26534) \leq n \ln(1.01837)$$ $$n \geq \frac{\ln(1.26534)}{\ln(1.01837)} = \frac{0.235}{0.0182} = 12.91$$ 8. **Interpret $n$:** $n$ is the number of 6-month periods. Since $n$ must be an integer and we round up to the next period: $$n = 13 \text{ periods}$$ 9. **Convert to years and months:** Each period = 6 months, so: $$13 \times 6 = 78 \text{ months} = 6 \text{ years and } 6 \text{ months}$$ **Final answer:** It will take 6 years and 6 months to save at least 13000 under the given conditions.