1. **Stating the problem:** We need to find the value of the sum $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots + \frac{1}{\sqrt{35} + \sqrt{36}}$$ 2. **Formula and approach:** To simplify each term, we use the conjugate to rationalize the denominator: $$\frac{1}{\sqrt{n} + \sqrt{n+1}} \times \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2}$$ 3. **Simplify the denominator:** $$= \frac{\sqrt{n+1} - \sqrt{n}}{n+1 - n} = \sqrt{n+1} - \sqrt{n}$$ 4. **Rewrite the sum:** $$\sum_{n=1}^{35} \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sum_{n=1}^{35} (\sqrt{n+1} - \sqrt{n})$$ 5. **Telescoping series:** Most terms cancel out: $$= (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{36} - \sqrt{35})$$ 6. **After cancellation:** $$= \cancel{\sqrt{2}} - \sqrt{1} + \cancel{\sqrt{3}} - \cancel{\sqrt{2}} + \cdots + \sqrt{36} - \cancel{\sqrt{35}} = \sqrt{36} - \sqrt{1}$$ 7. **Calculate the final value:** $$= 6 - 1 = 5$$ **Final answer:** $$5$$