## 4. Magic Table

In each cell of the table in the picture, one of the numbers $11, 22, 33, 44, 55$, 66, 77, 88, and 99 is written. The sums of the three numbers in the cells of each row and each column are equal. If the number 33 is written in the center cell of the table, what is the sum of the numbers written in the four colored cells?

![](https://cdn.mathpix.com/cropped/2024_05_30_43f5242198ec4dde97b4g-3.jpg?height=265&width=266&top_left_y=1044&top_left_x=1586)

Result: $\quad 198$
## Solution, First Method

Since $11+22+33+44+55+66+77+88+99=\cdots=495$, the sum of the numbers in each row and each column must be 495/3=165.

Consider the numbers A, B, C, and D (in the figure):

![](https://cdn.mathpix.com/cropped/2024_05_30_43f5242198ec4dde97b4g-3.jpg?height=228&width=237&top_left_y=1639&top_left_x=972)

Since $A+33+B=165$ and $C+3+D=165$, it must be that $A+B=132$ and $C+D=132$. Since $A+B+C+D+33=132+132+33=297$, we conclude that the sum of the remaining four numbers is $495-297=198$.

## Solution, Second Method

Since $11+22+33+44+55+66+77+88+99=\cdots=495$, the sum of the numbers in each row and each column must be 495/3=165.

Consider the numbers A, B, C, and D (in the figure):

![](https://cdn.mathpix.com/cropped/2024_05_30_43f5242198ec4dde97b4g-3.jpg?height=217&width=240&top_left_y=2367&top_left_x=908)

We need to determine which pair of numbers can be in the second row. Let's assume $A < B$. Since $A+33+B=165$, it must be that $A+B=132$. The largest value that $B$ can have is 99, so $A$ cannot be 11 or 22 because in that case, $B$ would need to be greater than 99. The number 33 is in the middle, so the smallest possible value for $A$ is 44. In that case, $B=132-44=88$.

Furthermore, $A$ can be 55. In that case, $B=132-55=77$. If $A$ were 66, then $B$ would also need to be 66, which is not possible because all numbers in the table must be different.

Therefore, the only two pairs of numbers that are possible are 44 and 88, and 55 and 77. Since the same applies to the numbers C and D, one of these pairs will be A and B, and the other will be C and D. The remaining numbers, which are 11, 22, 66, and 99, will be in the 4 colored cells. Their sum is 198.