\( \begin{array}{lllll}\text { (A) } 2 & \text { (B) } 28 & \text { (C) } 24 & \text { (D) } 22 & \text { (E) } 21\end{array} \) 10. How many positive integers less than 20000 are both perfect squares and perfect cubes? \( \begin{array}{lllll}\text { (A) } 4 & \text { (B) } 5 & \text { (C) } 6 & \text { (D) } 7 & \text { (E) } 8\end{array} \) 11. You are given that the real number \( x>0 \) satisfies \( (2 x)^{20}=5^{10} x^{10} \). Find the exact value of \( x \)

To tackle the first question about positive integers less than 20,000 that are both perfect squares and perfect cubes, we can think about perfect sixth powers. A number that is both a perfect square and a perfect cube is a perfect sixth power (since the least common multiple of 2 and 3 is 6). The sixth powers less than 20,000 are \( 1^6, 2^6, 3^6, 4^6\), and \(5^6\), which gives us \(1, 64, 729, 4096, 15625\). That makes 5 positive integers! Now, moving on to the second question, let's simplify the equation \((2x)^{20} = 5^{10} x^{10}\). We can rewrite it as \(2^{20} x^{20} = 5^{10} x^{10}\). Dividing both sides by \(x^{10}\) (noting that \(x > 0\)), we have \(2^{20} x^{10} = 5^{10}\). To isolate \(x^{10}\), divide both sides by \(2^{20}\), getting \(x^{10} = \frac{5^{10}}{2^{20}} = \left(\frac{5}{4}\right)^{10}\). Taking the tenth root gives us \(x = \frac{5}{4}\). So the answers are 5 for the first question and \( \frac{5}{4} \) for the second!