## Task 6 - 280736

On a circle, the natural numbers from 1 to 1000 are arranged in sequence. Starting with the number 1, every fifteenth number is marked, i.e., the numbers $1, 16, 31, 46, \ldots$ and so on are marked.

This process of counting and marking every fifteenth number is continued in a circular manner, meaning that after the number 1000, the number 1 follows again. Even numbers that are already marked are always taken into account during the counting. The process is only stopped when only numbers that are already marked would be reached for further marking.

Determine the number of all those numbers on the circle that remain unmarked!
In the first round, exactly those numbers are marked which, when divided by 15, leave a remainder of 1. The last of these numbers is 991.

In the second round, the first marked number is $991 + 15 - 1000 = 6$; consequently, in the second round, exactly those numbers are marked which, when divided by 15, leave a remainder of 6. The last of these numbers is 996.

Accordingly, in the third round, the first marked number is $996 + 15 - 1000 = 11$, then all numbers that, when divided by 15, leave a remainder of 11, and the last of these numbers is 986.

A further continuation would result in the number $986 + 15 - 1000 = 1$ and thus only already marked numbers, so the process is completed.

Thus, exactly those numbers are marked which, when divided by 15, leave a remainder of 1, 6, or 11. These are exactly the numbers that, when divided by 5, leave a remainder of 1, i.e., exactly the numbers $1, 1 \cdot 5 + 1 = 6, 2 \cdot 5 + 1 = 11, 3 \cdot 5 + 1 = 16, \ldots, 199 \cdot 5 + 1 = 996$.

Their number (as this enumeration shows, obtained as the number of numbers 0, 1, 2, 3, ... 199) is 200. Thus, exactly $1000 - 200 = 800$ numbers remain unmarked.

Solutions of the third round 1989 taken from [5]

### 4.31 29th Olympiad 1989

### 4.31.1 First Round 1989, Class 7