Solution

Let $A$ and $B$ be the numbers placed in the shaded cells as indicated in the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-34.jpg?height=153&width=426&top_left_y=1071&top_left_x=887)

Observe that, by the rules, $B$ must be greater than all the other numbers in its row and also greater than $A$. But since $A$ is greater than all the other numbers in its row, we conclude that $B$ must be greater than all the other numbers on the board. Therefore, $B=12$.

a) The largest value that $A$ can take is 11. The largest sum is then $A+B=11+12=$ 23. The following figure shows one way to achieve this sum:

| 1 | 3 | 5 | 7 | 9 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 4 | 6 | 8 | 10 | 12 |

Note: This is, in fact, the only way in which the sum is equal to 23.

b) Since $A$ must be greater than the other 5 numbers in its row, the minimum value that $A$ can take is 6. Therefore, the smallest value that the sum can take is $A+B=$ $6+12=18$. The following figure shows one way to achieve this sum:

| 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 7 | 8 | 9 | 10 | 11 | 12 |

Note: This is, in fact, the only way in which the sum is equal to 18.
Pedro has a $6 \times 2$ board, with the two rightmost cells painted gray as illustrated below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-11.jpg?height=222&width=616&top_left_y=1882&top_left_x=797)

He must fill all the cells of his board with the numbers from 1 to 12 such that:

- in each row, the 6 numbers, read from left to right, are in increasing order; and
- in each column, the number on top is less than the number below.
a) Show that the largest sum Pedro can achieve in the gray cells is 23. Also show how he can achieve this sum.

b) Show that the smallest sum Pedro can achieve in the gray cells is 18. Also show how he can achieve this sum.

#