## 9. Maximum Sum

In one of the fields in the image, the number 540 is written, and the other fields need to be filled with natural numbers greater than 1 such that the product of the numbers in two adjacent fields is equal to the number in the field below them. What is the maximum possible sum of these six numbers?

![](https://cdn.mathpix.com/cropped/2024_05_30_bfe8eca8742272ceb120g-6.jpg?height=452&width=451&top_left_y=1453&top_left_x=1405)

## Result: 686
Solution. The number 540 can be represented as the product of two factors greater than 1 in several ways, but if we decompose it, for example, as $540=2 \cdot 270$, and write these numbers in the middle row, we won't be able to fill the top row because we cannot use the factor 1. This actually means that none of the numbers in the middle row can be a prime number, so the number 540 must be represented as the product of two composite numbers. Additionally, these numbers must not be relatively prime, because in that case, we would have to write the number 1 in the middle cell of the top row. Therefore, we consider these possibilities:

$$
540=6 \cdot 90=9 \cdot 60=10 \cdot 54=12 \cdot 45=15 \cdot 36=18 \cdot 30
$$

In the upper middle cell, any common factor of these numbers (greater than 1) can be written, so we have the following possibilities:
![](https://cdn.mathpix.com/cropped/2024_05_30_bfe8eca8742272ceb120g-7.jpg?height=1674&width=1686&top_left_y=150&top_left_x=82)

The largest possible sum of all 6 numbers is 686, which is achieved in the first case $(45+2+3+90+6+ 540 )$.

| 15 | 11 | 37 | 11 | 15 |
| :--- | :--- | :--- | :--- | :--- |
| 11 | 21 | 40 | 21 | 11 |
| 37 | 40 | 26 | 40 | 37 |
| 11 | 21 | 40 | 21 | 11 |
| 15 | 11 | 37 | 11 | 15 |