1. **State the problem:** Find the values of $n$ such that $$\frac{10^{4n} \times 2^{3(n^2 - 5n)} \times 5^{2(1 - 2n)}}{20^2} = 1$$ 2. **Rewrite the denominator:** Note that $20 = 2 \times 10$, so $$20^2 = (2 \times 10)^2 = 2^2 \times 10^2$$ 3. **Substitute and rewrite the equation:** $$\frac{10^{4n} \times 2^{3(n^2 - 5n)} \times 5^{2(1 - 2n)}}{2^2 \times 10^2} = 1$$ 4. **Express all terms with prime bases 2 and 5:** Recall $10 = 2 \times 5$, so $$10^{4n} = (2 \times 5)^{4n} = 2^{4n} \times 5^{4n}$$ Substitute into numerator: $$2^{4n} \times 5^{4n} \times 2^{3(n^2 - 5n)} \times 5^{2(1 - 2n)} = 2^{4n + 3(n^2 - 5n)} \times 5^{4n + 2(1 - 2n)}$$ 5. **Combine powers of 2 and 5:** For base 2: $$4n + 3(n^2 - 5n) = 4n + 3n^2 - 15n = 3n^2 - 11n$$ For base 5: $$4n + 2(1 - 2n) = 4n + 2 - 4n = 2$$ 6. **Rewrite the entire fraction:** $$\frac{2^{3n^2 - 11n} \times 5^2}{2^2 \times 10^2} = 1$$ Recall $10^2 = (2 \times 5)^2 = 2^2 \times 5^2$, so denominator is $$2^2 \times 2^2 \times 5^2 = 2^{4} \times 5^{2}$$ 7. **Simplify the fraction:** $$\frac{2^{3n^2 - 11n} \times 5^2}{2^{4} \times 5^{2}} = 2^{3n^2 - 11n - 4} \times 5^{2 - 2} = 2^{3n^2 - 11n - 4} \times 5^{0} = 2^{3n^2 - 11n - 4}$$ 8. **Set the expression equal to 1:** Since $5^0 = 1$, the equation reduces to $$2^{3n^2 - 11n - 4} = 1$$ 9. **Solve for the exponent:** Because $2^x = 1$ only when $x = 0$, we have $$3n^2 - 11n - 4 = 0$$ 10. **Solve the quadratic equation:** Use the quadratic formula: $$n = \frac{11 \pm \sqrt{(-11)^2 - 4 \times 3 \times (-4)}}{2 \times 3} = \frac{11 \pm \sqrt{121 + 48}}{6} = \frac{11 \pm \sqrt{169}}{6} = \frac{11 \pm 13}{6}$$ 11. **Find the two roots:** - $$n = \frac{11 + 13}{6} = \frac{24}{6} = 4$$ - $$n = \frac{11 - 13}{6} = \frac{-2}{6} = -\frac{1}{3}$$ **Final answer:** $$\boxed{n = 4 \text{ or } n = -\frac{1}{3}}$$