## Task 4 - 130524

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0064.jpg?height=437&width=554&top_left_y=610&top_left_x=751)

In the Centrum department store, equal-sized cans have been stacked to form a "pyramid" for decorative purposes.

In each layer, the cans are arranged "in a triangle" as shown in the illustration.

The number of cans, denoted by $k$, along each "side of the triangle" in the bottom layer is 9. In each subsequent layer, the corresponding number $k$ is 1 less than in the immediately underlying layer. The top layer consists of a single can.

Determine the total number of cans contained in the "pyramid"!
The bottom layer consists of 9 rows, with the first row containing exactly 1 can and each subsequent row containing exactly one more can than the row immediately preceding it. The ninth row thus contains exactly 9 cans. Consequently, the total number of cans in this layer is equal to the sum of the natural numbers from 1 to 9, which is 45.

The layer immediately above contains exactly one row less, namely the row with 9 cans. The same applies to all other layers. Thus, we obtain:

| First layer: | 45 |
| :--- | :--- |
| second layer: | $36=45-9$ |
| third layer: | $28=36-8$ |
| fourth layer: | $21=28-7$ |
| fifth layer: | $15=21-6$ |
| sixth layer: | $10=15-5$ |
| seventh layer: | $6=10-4$ |
| eighth layer: | $3=6-3$ |
| ninth layer: | $1=3-2$ |
| in total: | 165 |

A total of 165 cans were used to build the "pyramid."

Solutions of the second round 1973 taken from [5]

### 2.15 XIV. Olympiad 1974

### 2.15.1 First Round 1974, Class 5