2002 AMC 10B Problem 14

Since $25 = 5^2$ and $64 = 2^6,$ we have $25^{64}\cdot 64^{25} = 5^{128}\cdot 2^{150},$ so $$N = \sqrt{5^{128}\cdot 2^{150}} = 5^{64}\cdot 2^{75}.$$ Writing $2^{75} = 2^{64}\cdot 2^{11},$ we get $N = (5\cdot 2)^{64}\cdot 2^{11} = 10^{64}\cdot 2048.$ So $N$ is $2048$ followed by $64$ zeros, and its digit sum is $2 + 0 + 4 + 8 = 14.$ Thus, the correct answer is B.