1. **State the problem:** Find the smallest number $n$ such that $$169^{80} \leq n^{120}.$$\n\n2. **Use the inequality and properties of exponents:** We want to solve for $n$ in $$169^{80} \leq n^{120}.$$\n\n3. **Rewrite the inequality:** Taking the 120th root on both sides (which preserves inequality since 120 is positive), we get $$\sqrt[120]{169^{80}} \leq n.$$\n\n4. **Simplify the root:** Using the property $$\sqrt[m]{a^k} = a^{\frac{k}{m}},$$ we have $$n \geq 169^{\frac{80}{120}} = 169^{\frac{2}{3}}.$$\n\n5. **Express 169 as a power:** Since $$169 = 13^2,$$ substitute to get $$n \geq (13^2)^{\frac{2}{3}} = 13^{\frac{4}{3}}.$$\n\n6. **Interpret the exponent:** $$13^{\frac{4}{3}} = 13^{1 + \frac{1}{3}} = 13 \times 13^{\frac{1}{3}} = 13 \times \sqrt[3]{13}.$$\n\n7. **Final answer:** The smallest number $n$ satisfying the inequality is $$n = 13 \times \sqrt[3]{13}.$$