2004 AMC 10B Problem 4

The six faces multiply to $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 720$. If the hidden bottom face shows $h$, then the visible product is $P = 720/h$. As $h$ runs through $1, \dots, 6$, the possible values of $P$ are
$$720,\ 360,\ 240,\ 180,\ 144,\ 120.$$

A number is certain to divide $P$ exactly when it divides all six of these, so we want their greatest common divisor. Dividing $720$ by each of $1, \dots, 6$ and taking the gcd is the same as dividing $720$ by the least common multiple of $1, \dots, 6$:
$$\gcd\!\left(\tfrac{720}{1}, \dots, \tfrac{720}{6}\right) = \frac{720}{\operatorname{lcm}(1,2,3,4,5,6)} = \frac{720}{60} = 12.$$

Concretely: hiding $5$ kills the factor $5$; hiding $4$ leaves only $2^2$ among the powers of $2$ (from $2$ and $6$); hiding $3$ or $6$ leaves a single factor of $3$. What survives every case is $2^2 \cdot 3 = 12$.

The answer is $\boxed{\textbf{(B)}\ 12}$.

"Certain to divide" means "divides in the worst case," so for each prime you look for the roll that removes the most copies of that prime. Rewriting $P$ as $720/h$ turns the six cases into a single gcd computation, and $\gcd(N/a_i) = N/\operatorname{lcm}(a_i)$ whenever all $a_i$ divide $N$.

Check a few values directly: $P = 180$ (hide $4$) and $P = 120$ (hide $6$) already have $\gcd(180, 120) = 60$, and adding $P = 144$ (hide $5$) gives $\gcd(60, 144) = 12$. Since $12$ also divides $720$, $360$, and $240$, the answer is $12$.

• Reasoning from one typical roll (hiding 1 leaves 720) instead of the worst case for each prime: hiding 4 and hiding 6 are what remove the extra 2s and 3s.
• Assuming a factor of $5$ is guaranteed because $5$ is "usually" visible, which leads to $60$ or $144$-style overcounts.