2005 AMC 10A Problem 15

Factor each factorial: $3! = 2 \cdot 3$, $5! = 2^3 \cdot 3 \cdot 5$, $7! = 2^4 \cdot 3^2 \cdot 5 \cdot 7$. Multiplying and adding exponents,
$$3! \cdot 5! \cdot 7! = 2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7.$$

A divisor $2^a 3^b 5^c 7^d$ is a perfect cube exactly when every exponent is a multiple of $3$, and it divides the product when $a \leq 8$, $b \leq 4$, $c \leq 2$, $d \leq 1$. The options:

• $a \in \{0, 3, 6\}$: $3$ choices,
• $b \in \{0, 3\}$: $2$ choices,
• $c = 0$ and $d = 0$: $1$ choice each.

By the multiplication principle there are $3 \cdot 2 \cdot 1 \cdot 1 = 6$ cube divisors, namely $1, 8, 27, 64, 216, 1728$.

The answer is $\boxed{\textbf{(E)}\ 6}$.

Every question about divisors reduces to exponents in the prime factorization. "Cube divisor" means each exponent is a multiple of $3$ not exceeding the available supply, so count the allowed exponents per prime and multiply. The same template counts square divisors (multiples of $2$) or divisors of any shape.

• Miscounting the exponent of 2 (it is 1 + 3 + 4 = 8, allowing 2^6) or forgetting that 1 = 1^3 counts as a cube.
• Listing cubes by inspection ($1, 8, 27, 64$) and missing the mixed ones $6^3 = 216$ and $12^3 = 1728$.