A standard six-sided die is rolled, and  $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$?
The possible values of $P$ depend on which face is hidden, so there are six cases to consider. If the hidden face is 1, then $P$ is the product of 2, 3, 4, 5, and 6, which is 720. If the hidden face is 2, then $P$ is the product of 1, 3, 4, 5, and 6, which is 360. If the hidden face is 3, then $P$ is the product of 1, 2, 4, 5, and 6, which is 240. If the hidden face is 4, then $P$ is the product of 1, 2, 3, 5, and 6, which is 180. If the hidden face is 5, then $P$ is the product of 1, 2, 3, 4, and 6, which is 144. If the hidden face is 6, then $P$ is the product of 1, 2, 3, 4, and 5, which is 120. The largest number that is certain to divide $P$ is the greatest common factor of all these possible values. To find the greatest common factor, I can use prime factorization and look for the common prime factors and their lowest powers. 720 = 2^4 * 3^2 * 5^1, 360 = 2^3 * 3^2 * 5^1, 240 = 2^4 * 3^1 * 5^1, 180 = 2^2 * 3^2 * 5^1, 144 = 2^4 * 3^2, 120 = 2^3 * 3^1 * 5^1. The common prime factors are 2, 3, and 5, and their lowest powers are 2^2, 3^1, and 5^0, respectively. The greatest common factor is 2^2 * 3^1 * 5^0 = 12. Therefore, the largest number that is certain to divide $P$ is 12. 