1. **Stating the problem:** We are given the equation $$\frac{1}{a} + \frac{1}{b} = \frac{1}{13}$$ and need to find the value of $$a + b$$. 2. **Rewrite the equation:** Combine the fractions on the left-hand side over a common denominator: $$\frac{b}{ab} + \frac{a}{ab} = \frac{1}{13}$$ which simplifies to $$\frac{a + b}{ab} = \frac{1}{13}$$. 3. **Cross-multiply:** Multiply both sides by $$13ab$$ to clear denominators: $$13(a + b) = ab$$. 4. **Rewrite the equation:** $$ab = 13(a + b)$$. 5. **Express in terms of $$a$$ and $$b$$:** Rearranged as $$ab - 13a - 13b = 0$$. 6. **Add $$169$$ to both sides to complete the rectangle:** $$ab - 13a - 13b + 169 = 169$$. 7. **Factor the left side:** $$(a - 13)(b - 13) = 169$$. 8. **Analyze the factorization:** Since $$169 = 13^2$$, possible integer factor pairs are: $$(1, 169), (13, 13), (169, 1), (-1, -169), (-13, -13), (-169, -1)$$. 9. **Find $$a$$ and $$b$$ from each pair:** For $$(a - 13)(b - 13) = 169$$, if $$a - 13 = 1$$ and $$b - 13 = 169$$, then $$a = 14$$ and $$b = 182$$. Similarly, if $$a - 13 = 13$$ and $$b - 13 = 13$$, then $$a = 26$$ and $$b = 26$$. 10. **Calculate $$a + b$$ for these pairs:** - For $$a=14, b=182$$, $$a + b = 196$$. - For $$a=26, b=26$$, $$a + b = 52$$. 11. **Verify which pair satisfies the original equation:** - For $$a=14, b=182$$: $$\frac{1}{14} + \frac{1}{182} = \frac{13}{182} + \frac{1}{182} = \frac{14}{182} = \frac{7}{91} \neq \frac{1}{13}$$. - For $$a=26, b=26$$: $$\frac{1}{26} + \frac{1}{26} = \frac{2}{26} = \frac{1}{13}$$ which matches the original equation. 12. **Final answer:** $$a + b = 26 + 26 = 52$$.