2023 AMC 10A Problem 5

The number is far too big to multiply out, so rewrite it using primes. Every $2$ paired with a $5$ makes a $10$, and $10$s only add zeros.

1. $8 = 2 \times 2 \times 2$, so $8^5 = 2^{15}$ (fifteen $2$s).
2. $15 = 3 \times 5$, so $15^5 = 3^5 \times 5^5$.
3. Put everything together:
$$8^5 \cdot 5^{10} \cdot 15^5 = 2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5 = 2^{15} \cdot 5^{15} \cdot 3^5.$$
4. There are fifteen $2$s and fifteen $5$s. Pair them up: $2^{15} \cdot 5^{15} = 10^{15}$.
5. The leftover part is $3^5 = 243$.
6. So the number is $243 \times 10^{15}$, which is $243$ followed by fifteen zeros.
7. Count the digits: $3$ digits in $243$ plus $15$ zeros gives $18$ digits.

The answer is $\boxed{\textbf{(E)}\ 18}$.

To count the digits of a big product, look for $2$s and $5$s: each pair makes a $10$ and just adds a zero at the end. Whatever is left over is usually a small number whose digits you can count directly. Breaking every number into primes first is what makes the pairs easy to see.