5-7. On the faces of a die, the numbers $6,7,8,9,10,11$ are written. The die was rolled twice. The first time, the sum of the numbers on the four "vertical" (that is, excluding the bottom and top) faces was 33, and the second time - 35. What number can be written on the face opposite the face with the number 7? Find all possible options.

![](https://cdn.mathpix.com/cropped/2024_05_06_d9f67fd4a24dc8d44db8g-07.jpg?height=531&width=534&top_left_y=714&top_left_x=767)
Answer: 9 or 11.

Solution. The total sum of the numbers on the faces is $6+7+8+9+10+11=51$. Since the sum of the numbers on four faces the first time is 33, the sum of the numbers on the two remaining faces is $51-33=18$. Similarly, the sum of the numbers on two other opposite faces is $51-35=16$. Then, the sum on the third pair of opposite faces is $51-16-18=17$. Let's consider the three cases in which the number 7 falls into the sum (16, 17, or 18):

- $7+9=16$, this is possible if the pairs are $6+11=17$ and $8+10=18$.
- $7+10=17$, this is not possible, as the number 6 cannot be part of the sum 16 (since the number 10 is already used), and it cannot be part of the sum 18 (since there is no number 12).
- $7+11=18$, this is possible if the pairs are $6+10=16$ and $8+9=17$.