# 8. Variant 1.

101 natural numbers are written in a circle. It is known that among any 5 consecutive numbers, there will be at least two even numbers. What is the minimum number of even numbers that can be among the written numbers?
Answer: 41.

Solution. Consider any 5 consecutive numbers. Among them, there is an even number. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there are at least two even numbers. Thus, the total number of even numbers is no less than $1+2 \cdot 20=41$. Such a situation is possible. Number the numbers in a circle. The even numbers can be those with numbers $1,3,6$, $8,11, \ldots, 98,101$ (the number with number 1 is even, and in each set of 5 numbers with numbers $5 \mathrm{k}+2$, $5 \mathrm{k}+3,5 \mathrm{k}+4,5 \mathrm{k}+5,5 \mathrm{k}+6$, the numbers with numbers $5 \mathrm{k}+3$ and $5 \mathrm{k}+6$ are even).