2002 AMC 10B Problem 14

The digit sum of a giant number is only easy if the number turns out to be simple, like a small number followed by zeros. A $2$ times a $5$ makes a $10$, which just adds a zero. So break everything into $2$s and $5$s.

1. Write each base as a prime power: $25 = 5^2$ and $64 = 2^6$. So
$$25^{64} = 5^{128}, \qquad 64^{25} = 2^{150}.$$
2. $N$ is the square root. Taking a square root halves each exponent (because $5^{64} \cdot 5^{64} = 5^{128}$):
$$N = 5^{64} \cdot 2^{75}.$$
3. There are $64$ fives and $75$ twos. Pair every five with a two to make $64$ tens. That uses $64$ twos, leaving $75 - 64 = 11$ twos:
$$N = 10^{64} \cdot 2^{11}.$$
4. $2^{11} = 2048$. Multiplying by $10^{64}$ writes $64$ zeros after it, so $N$ is $2048$ followed by $64$ zeros.
5. Digit sum: $2 + 0 + 4 + 8 = 14$.

The answer is $\boxed{\textbf{(B)}\ 14}$.

Powers of $10$ only add zeros, and zeros add nothing to a digit sum. So whenever a huge number is made of $2$s and $5$s, pair them up into tens and look only at what is left over. That small leftover number decides the digits.

Write everything in primes: $25^{64} = 5^{128}$ and $64^{25} = (2^6)^{25} = 2^{150}$. Taking the square root halves each exponent:
$$N = 5^{64}\cdot 2^{75}.$$

Every $5$ can be matched with a $2$ to make a $10$; there are $64$ fives and $75$ twos, so $64$ pairs form and $11$ twos are left over:
$$N = (2\cdot 5)^{64}\cdot 2^{11} = 2048 \cdot 10^{64}.$$

Multiplying by $10^{64}$ just appends $64$ zeros, so the nonzero digits of $N$ are those of $2048$, with sum $2 + 0 + 4 + 8 = 14$.

The answer is $\boxed{\textbf{(B)}\ 14}$.