05/22/2020 (2011: National Team Round, Problem 4)

Q: Six standard dice are stacked on the floor as shown below. On each die the sum of the opposite faces is 7. What is the maximum possible sum of the 21 visible faces?

A: 89

There are a total of 36 sides on the 6 dice. That means that the sum of the total numbers on the cubes is 7 * 3 * 6 = 126.

The top die has one side covered. That means that the total sum goes down by, at least, 1. That makes the sum as of now 126 - 1 = 125

The die below the topmost one has 3 sides, 2 of which are opposing, covered. That means that the sum of the numbers goes down, at the least, by 8 (7 + 1). That makes the sum as of now 125 - 8 = 117.

The die below the one below the topmost one has 3 sides, 2 of which are opposing, covered. That means that the sum of the numbers goes down, at the least, by 8 (7 + 1). That makes the sum as of now 117 - 8 = 109.

The die next to that one has 4 sides, 2 pairs of opposite sides, covered. That means the sum of the numbers goes down by 14 (7 * 2). That makes the sum as of now 109 - 14 = 95.

The die above it has 2 sides, neither opposing, covered. That means the sum of the numbers goes down by, at the least, 3 (1 + 2). That makes the sum as of now 95 - 3 = 92.

The last remaining die has 2 sides, neither opposing, covered. That means the sum of the numbers goes down by, at the least, 3 (1 + 2). That makes the sum as of now 92 - 3 = 89