2002 AMC 10B Problem 14

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 $\boxed{\textbf{(B)}\ 14}$.

Digit sums of huge numbers are tractable only when the number has a simple decimal form. Powers of $2$ and $5$ combine into powers of $10$, which contribute nothing to the digit sum, so the whole problem reduces to the small leftover factor. Always pair off $2$s and $5$s before thinking about digits.

• Leaving the wrong prime unpaired (writing 10^75 5^{-11} or 10^64 5^11) or forgetting to halve the exponents when taking the square root.
• Miscomputing $2^{11}$ (for instance as $1024$ or $4096$), which changes the digit sum to $7$ or $19$.