## 6. Three Cubes

| 45 | 56 |  |  |
| :---: | :---: | :---: | :---: |
|  | 92 | 108 |  |
|  |  | 68 | 75 |

Three identical cubes have a net as shown in the figure. The cubes are placed on a table, one on top of the other. What is the smallest possible sum of the numbers on all visible faces of the cubes placed on the table?

![](https://cdn.mathpix.com/cropped/2024_05_30_adbfbded7af8db94c594g-3.jpg?height=323&width=282&top_left_y=2437&top_left_x=1618)
## Solution.

When such three cubes are placed on a table, one on top of the other, 5 sides are not visible, and the remaining 13 are visible. To determine the smallest possible sum of the numbers on all visible sides of the cubes, we need to determine the largest possible sum of the numbers on the 5 sides that are not visible.

When a cube is formed from the given net, the pairs of numbers on opposite sides are 45 and 108, 92 and 75, 56 and 68. The largest possible sum of such a pair is $92+75=167$. The bottom two cubes have non-visible sides with numbers 92 and 75, and the top cube has a non-visible side with the number 108.

The sum of all hidden numbers is $167+167+108=442$.

The sum of the numbers on each cube is $45+56+92+108+68+75=444$, so the sum for three cubes is 3 . $444=1332$.

Subtract the sum of the hidden numbers from the sum of all numbers, $1332-442=890$.

The smallest possible sum of the numbers on all visible sides of the cubes is 890.