Problem 4. On a board, eight numbers equal to 0 are written, one after another. We call an operation the modification of four of the eight numbers as follows: two numbers increase by 3, one number increases by 2, and the fourth number increases by 1.

a) What is the minimum number of operations we need to perform to obtain eight consecutive natural numbers on the board.

b) Is it possible that, after a number of operations, all the numbers written on the board are equal to 2015?

c) Is it possible that, after a sequence of operations, the product of the numbers on the board is 2145?
## Solution.

a) With each operation, the sum of the numbers increases by 9. This means that after $k$ operations, the sum of the numbers on the board will be $9 \cdot k$. The sum of the smallest eight consecutive natural numbers is $0+1+2+\ldots+7=28$, which is not divisible by 9, while $1+2+3+\ldots+8=36$. Since $9 \cdot 4=36$, we deduce that the minimum number of operations required to obtain eight consecutive numbers is 4. Here is one such possibility:

| Initial | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Operation 1 | 1 | 2 |  |  |  |  | 3 | 3 |
| Operation 2 |  |  | 2 | 1 |  |  | 3 | 3 |
| Operation 3 |  |  |  |  | 3 | 3 | 1 | 2 |
| Operation 4 |  |  | 1 | 3 | 2 | 3 |  |  |
| Total | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

....................................................................................................... 2 p

b) As observed in part a), after $k$ operations, the sum of the 8 numbers is $9 \cdot k$. Assuming that after a certain number of operations all 8 numbers are equal to 2015, then their sum is $8 \cdot 2015$. Since $8 \cdot 2015$ is not divisible by 9, we conclude that it is not possible for all the numbers on the board to be equal to 2015 after any number of operations .................................................... 2p

c) We have $2145=3 \cdot 5 \cdot 11 \cdot 13$. If the numbers on the board are $3,5,11,13,1,1,1,1$, then $3+5+11+13+1+1+1+1=36$, which means that 4 operations are required to reach these eight numbers. Since we cannot obtain 13 from 4 operations, we deduce that this variant is not possible.

We cannot have 6 or more numbers equal to 1 on the board, since in any selection, three of the numbers are greater than 1. If there are 5 numbers equal to 1 on the board, it means that 5 operations have been performed. This results in $a+b+c+1+1+1+1+1=45$, where $a$ is the product of two of the numbers $3,5,11$ or 13, and $b$ and $c$ are the remaining two numbers. In none of these cases is the sum $a+b+c$ equal to 40. Therefore, this variant is also not possible. In conclusion, it is not possible for the product of the numbers on the board to be 2145 after any sequence of operations.