1. **State the problem:** You have a mortgage loan of $316,000 with an interest rate of 4.1% per year, paid monthly over 30 years. Monthly payments are $1,526.91. We want to find the remaining balance after 20 years of payments. 2. **Formula used:** The remaining balance on a mortgage after $n$ payments is given by the formula: $$B = P \left(1 + \frac{r}{12}\right)^{N} - \frac{M}{\frac{r}{12}} \left[ \left(1 + \frac{r}{12}\right)^{N} - 1 \right]$$ where: - $B$ is the balance after $n$ payments, - $P$ is the original loan amount, - $r$ is the annual interest rate (decimal), - $M$ is the monthly payment, - $N$ is the number of remaining payments. 3. **Identify values:** - $P = 316000$ - $r = 0.041$ - Total payments over 30 years: $30 \times 12 = 360$ - Payments made: $20 \times 12 = 240$ - Remaining payments: $N = 360 - 240 = 120$ - Monthly payment: $M = 1526.91$ 4. **Calculate monthly interest rate:** $$i = \frac{r}{12} = \frac{0.041}{12} = 0.0034167$$ 5. **Calculate the balance after 20 years:** $$B = 316000 \times (1 + 0.0034167)^{120} - \frac{1526.91}{0.0034167} \times \left[(1 + 0.0034167)^{120} - 1\right]$$ 6. **Calculate powers:** $$ (1 + 0.0034167)^{120} = (1.0034167)^{120} \approx 1.491825 $$ 7. **Calculate each term:** $$316000 \times 1.491825 = 471,674.7$$ $$\frac{1526.91}{0.0034167} = 446,849.5$$ $$446,849.5 \times (1.491825 - 1) = 446,849.5 \times 0.491825 = 219,801.3$$ 8. **Calculate remaining balance:** $$B = 471,674.7 - 219,801.3 = 251,873.4$$ 9. **Round to nearest dollar:** $$\boxed{251,873}$$ So, the amount still owed after 20 years is $251,873.