1. **State the problem:** You have 15,000 euros now in the bank with an interest rate of 2.4% per year. We want to find out how much money should have been deposited 10 years ago to grow to 15,000 euros today. 2. **Formula used:** The formula for compound interest is $$A = P(1 + r)^t$$ where: - $A$ is the amount of money accumulated after $t$ years, including interest. - $P$ is the principal amount (the initial deposit). - $r$ is the annual interest rate (in decimal). - $t$ is the time the money is invested for in years. 3. **Rearrange the formula to find $P$:** $$P = \frac{A}{(1 + r)^t}$$ 4. **Plug in the values:** - $A = 15000$ - $r = 0.024$ (2.4% as a decimal) - $t = 10$ $$P = \frac{15000}{(1 + 0.024)^{10}}$$ 5. **Calculate the denominator:** $$1 + 0.024 = 1.024$$ $$1.024^{10} \approx 1.26824$$ 6. **Calculate $P$:** $$P = \frac{15000}{1.26824} \approx 11826.5$$ 7. **Round to the nearest integer:** $$P \approx 11827$$ **Final answer:** You should have deposited approximately **11827** euros 10 years ago to have 15000 euros now at 2.4% annual interest.