2005 AMC 10A Problem 15

Questions about divisors are easiest with prime factorizations, so factor everything first.

1. Factor each factorial.
- $3! = 6 = 2 \cdot 3$
- $5! = 120 = 2^3 \cdot 3 \cdot 5$
- $7! = 5040 = 2^4 \cdot 3^2 \cdot 5 \cdot 7$
2. Multiply by adding exponents. Twos: $1 + 3 + 4 = 8$. Threes: $1 + 1 + 2 = 4$. Fives: $1 + 1 = 2$. Sevens: $1$.
$$3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.$$
3. What does a cube divisor look like? A divisor uses each prime at most as many times as the product has it. It is a perfect cube when each prime appears $0, 3, 6, \ldots$ times.
4. Choices for each prime.
- Two: $0$, $3$ or $6$ times (not $9$, since only $8$ are available): $3$ choices.
- Three: $0$ or $3$ times: $2$ choices.
- Five: only $2$ available, so $0$ times: $1$ choice.
- Seven: only $1$ available, so $0$ times: $1$ choice.
5. Multiply the choices. $3 \times 2 \times 1 \times 1 = 6$.

The six cubes are $1, 8, 27, 64, 216, 1728$.

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

Every divisor is built by choosing how many of each prime to use, up to what is available. A perfect cube needs each count to be a multiple of $3$, and $0$ counts too. Count the allowed choices prime by prime and multiply. The same method counts square divisors (use multiples of $2$).

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 $\boxed{\textbf{(E)}\ 6}$.