1. **Problem statement:** Universal Robina Corporation wants the annual depreciation cost to never exceed 20% of the first cost for any year using the Sum of the Years' Digits (SYD) method, with no salvage value. We need to find the length of service $n$ in years. 2. **Formula for SYD depreciation:** The depreciation expense for year $k$ is given by: $$\text{Depreciation}_k = \frac{n - k + 1}{\frac{n(n+1)}{2}} \times \text{Cost}$$ where $n$ is the total service life in years, and $k$ is the year number. 3. **Important rule:** The maximum depreciation expense occurs in the first year ($k=1$), so: $$\text{Depreciation}_1 = \frac{n}{\frac{n(n+1)}{2}} \times \text{Cost} = \frac{2n}{n(n+1)} \times \text{Cost} = \frac{2}{n+1} \times \text{Cost}$$ 4. **Condition given:** The depreciation cost in any year should not exceed 20% of the first cost: $$\text{Depreciation}_1 \leq 0.20 \times \text{Cost}$$ Substitute the expression for $\text{Depreciation}_1$: $$\frac{2}{n+1} \times \text{Cost} \leq 0.20 \times \text{Cost}$$ 5. **Cancel Cost from both sides:** $$\cancel{\text{Cost}} \times \frac{2}{n+1} \leq 0.20 \times \cancel{\text{Cost}}$$ which simplifies to: $$\frac{2}{n+1} \leq 0.20$$ 6. **Solve for $n$:** Multiply both sides by $n+1$: $$2 \leq 0.20 (n+1)$$ Divide both sides by 0.20: $$\frac{2}{0.20} \leq n+1$$ $$10 \leq n+1$$ Subtract 1 from both sides: $$9 \leq n$$ 7. **Interpretation:** The length of service $n$ must be at least 9 years to ensure the annual depreciation cost does not exceed 20% of the first cost. **Final answer:** $$\boxed{n \geq 9 \text{ years}}$$