5. The numbers $1,2,3, \ldots, 999,1000$ are written one after another in this order on a circle. We cross out the number 1, and then every 15th number $(1,16,31, \ldots)$. When we complete one full circle, we count again, including the crossed-out numbers. How many numbers will remain uncrossed?
Solution. In the first round, we cross out the numbers 1, 16, 31, ..., 991 (all numbers of the form $15k + 1$), in the second round, we cross out the numbers $6, 21, 36, \ldots, 996$ (all numbers of the form $15k + 6$), in the third round, the numbers $11, 26, 41, \ldots, 986, 1$ (numbers of the form $15k + 11$), where $k=0, 1, \ldots, 66$. After the third round, only the numbers that have already been crossed out will be crossed out again. Thus, the numbers that are crossed out from 1 to 1000 are all those that give a remainder of 1 when divided by 5, i.e., they are of the form
$5m + 1, m \in \mathbb{N}_{0}$. There are 200 such numbers, so the number of uncrossed numbers will be $1000 - 200 = 800$ numbers.

## VIII Department - Ninth Grade