1. What is the minimum number of year-end deposits that have to be made before
the total value of the deposits is at least ten times greater than the value of a single
year-end deposit if the interest rate is 12.5%?

We are tasked with determining the minimum number of year-end deposits that must be made so that the total value of the deposits, compounded at an interest rate of 12.5%, is at least ten times greater than the value of a single year-end deposit.

### Steps:

1. **Formula for the Future Value of an Annuity**:
   The future value of a series of deposits (annuity) with interest compounded yearly is given by:

   \[
   S_n = D \times \frac{(1 + r)^n - 1}{r}
   \]

   Where:
   - \( S_n \) is the future value after \( n \) years,
   - \( D \) is the value of each year-end deposit,
   - \( r \) is the annual interest rate,
   - \( n \) is the number of deposits (years).

2. **Requirement**:
   We need the total future value \( S_n \) to be at least ten times a single deposit, i.e.,

   \[
   S_n \geq 10D
   \]

   Substituting the formula for \( S_n \):

   \[
   D \times \frac{(1 + r)^n - 1}{r} \geq 10D
   \]

   Dividing through by \( D \) (assuming \( D \neq 0 \)):

   \[
   \frac{(1 + r)^n - 1}{r} \geq 10
   \]

   Substituting \( r = 0.125 \) (since the interest rate is 12.5%):

   \[
   \frac{(1 + 0.125)^n - 1}{0.125} \geq 10
   \]

3. **Simplifying**:
   \[
   \frac{(1.125)^n - 1}{0.125} \geq 10
   \]

   Multiply both sides by 0.125:

   \[
   (1.125)^n - 1 \geq 1.25
   \]

   Adding 1 to both sides:

   \[
   (1.125)^n \geq 2.25
   \]

4. **Solving for \( n \)**:
   Taking the natural logarithm of both sides:

   \[
   \ln((1.125)^n) \geq \ln(2.25)
   \]

   Using the logarithmic property \( \ln(a^b) = b \ln(a) \):

   \[
   n \ln(1.125) \geq \ln(2.25)
   \]

   Now, calculate \( \ln(1.125) \) and \( \ln(2.25) \):

   \[
   n \times 0.117783 \geq 0.81093
   \]

   Solving for \( n \):

   \[
   n \geq \frac{0.81093}{0.117783} \approx 6.88
   \]

5. **Conclusion**:
   Since \( n \) must be an integer, we round up to the next whole number. Therefore, the minimum number of year-end deposits is:

   \[
   n = 7
   \]

### Answer:
The minimum number of year-end deposits is **7**.

---

Would you like further details or have any other questions?

Here are 5 related questions for further exploration:
1. How does the future value formula change if the deposits are made monthly instead of yearly?
2. What would be the minimum number of deposits if the interest rate were lower, say 5%?
3. How is the annuity formula affected if the interest rate compounds more frequently, like quarterly?
4. Can we derive the same result using a geometric series approach?
5. What if the goal were to triple the total value of deposits instead of making it 10 times larger?

**Tip**: When working with compound interest, using logarithms is often essential to solving for time or number of periods.

What is the minimum number of year-end deposits that have to be made before the total value of the deposits is at least ten times greater than the value of a single year-end deposit if the interest rate is 12.5%?

Discover how to calculate the minimum number of year-end deposits required to make the total value at least 10 times greater than a single deposit at an interest rate of 12.5%. This step-by-step guide explores the future value of an annuity, logarithmic properties, and compound interest calculations. Suitable for Grades 10-12.

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