2021 AMC Fall 10B Problem 14

Track only factors of $2$. Each die contributes $0$ factors of $2$ (a roll of $1, 3, 5$; probability $\frac12$), exactly $1$ factor (a roll of $2$ or $6$; probability $\frac13$), or $2$ factors (a roll of $4$; probability $\frac16$).

The product fails to be divisible by $4$ when the total number of factors of $2$ is $0$ or $1$.

• Zero factors: all six dice odd, probability $\left(\frac12\right)^6 = \frac{1}{64}$.
• Exactly one factor: one die shows $2$ or $6$ and the other five are odd. Choose which die ($6$ ways): $6 \cdot \frac13 \cdot \left(\frac12\right)^5 = \frac{6}{96} = \frac{1}{16} = \frac{4}{64}$.

The failing probability is $\frac{1}{64} + \frac{4}{64} = \frac{5}{64}$, so the product is divisible by $4$ with probability
$$1 - \frac{5}{64} = \frac{59}{64}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{59}{64}}$.

Divisibility of a product by $2^k$ depends only on the total exponent of $2$, so classify each die by how many 2s it contributes. The event "divisible by $4$" is large and messy, but its complement (total exponent at most $1$) has only two small cases.

• Treating a roll of 4 like a 2 or 6 (it already supplies two factors of 2), or forgetting the factor of 6 for which die is the lone even one.
• Computing "at least two even dice" instead, which ignores that a single $4$ suffices, and arriving at $\frac{57}{64}$.