8. Michael prints the result below twice on cardboard and makes two identical dice such that the visible dots are on the outside of the dice. He stacks them to form a tower. The front side of the bottom die shows 3 dots. The total number of dots on the two touching sides of the dice is 9. The total number of dots on the back of the tower is three times the total number of dots on the right side of the tower.

How many dots are on the side that touches the ground?
A) 1
B) 2
C) 4
D) 5
E) 6

![](https://cdn.mathpix.com/cropped/2024_04_17_4e6ce4fdc7c7bcbaf14eg-2.jpg?height=428&width=331&top_left_y=1679&top_left_x=1522)

## B-questions
A8. A) 1 On the bottom die, we see 3 eyes on the front, so there is a 4 on the back. On the right side of the bottom die, there can still be 1, 2, 5, or 6. With 5 or 6, the total number of eyes on the back must be at least 15 or 18, and that is not possible. If there is a 1 on the right side, then there is a 2 on the top (look at the picture), and thus there must be a 7 on the bottom of the top die. That is also not possible. Therefore, there is a 2 on the right side of the bottom die. On the side that touches the ground, there is thus a 1.

To verify, we look at the rest of the tower. On the top of the bottom die, we now find the 6, so there is a 3 on the bottom of the top die. There are then four ways in which we can place the top die. We know that the total number of eyes on the back of the tower is three times the total number of eyes on the right side of the tower. Testing gives that there is a 1 on the right side of the top die and a 5 on the back. Indeed, \(3 \cdot (2+1) = 4+5\).