\( \begin{array}{lllll}\text { (A) } 2 & \text { (B) } 28 & \text { (C) } 24 & \text { (D) } 22 & \text { (E) } 21\end{array} \) 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 \( x \)

For problem 10, a number that is both a perfect square and a perfect cube must be a perfect sixth power (since the least common multiple of 2 and 3 is 6). Let the number be of the form \[ n^6 < 20000. \] We test small positive integers: - For \( n=1 \): \( 1^6 = 1 \). - For \( n=2 \): \( 2^6 = 64 \). - For \( n=3 \): \( 3^6 = 729 \). - For \( n=4 \): \( 4^6 = 4096 \). - For \( n=5 \): \( 5^6 = 15625 \). - For \( n=6 \): \( 6^6 = 46656 \) (which is greater than 20000). Thus, the possible values of \( n \) are 1, 2, 3, 4, and 5. There are 5 such numbers. The answer to problem 10 is \(\textbf{(B) }5\). --- For problem 11, we start with the equation \[ (2x)^{20} = 5^{10} x^{10}. \] We rewrite the left-hand side using exponent rules: \[ (2x)^{20} = 2^{20} x^{20}. \] Thus, the equation becomes: \[ 2^{20} x^{20} = 5^{10} x^{10}. \] Since \( x>0 \), we can divide both sides by \( x^{10} \): \[ 2^{20} x^{10} = 5^{10}. \] Now solve for \( x^{10} \): \[ x^{10} = \frac{5^{10}}{2^{20}}. \] Notice that \( 2^{20} = (2^2)^{10} = 4^{10} \). Therefore, \[ x^{10} = \frac{5^{10}}{4^{10}} = \left(\frac{5}{4}\right)^{10}. \] Taking the tenth root of both sides, we obtain: \[ x = \frac{5}{4}. \] Thus, the exact value of \( x \) is \(\frac{5}{4}\).