Hi,

Let X = {1, 2,..., 10} =[10]. Suppose the elements of X are scattered at random around a circle. Show that there exists some string of three consecutive numbers whose sum is at least 18.

My attempt:

Let a1, a2, ..., a10 be the numbers.

There are 10 possible strings of 3 consecutive numbers, s1 = a1 + a2 + a3, s2 = a2 + a3 + a4, s3 = ... , s8 = a8 + a9 + a10, s9 = a9 + a10 + a1 and s10 = a10+a1+a2.

Assume each of these have magnitude less than 18. The sum of all these 10 numbers s1 + s2 + ... + s10 <= 10x17 = 170

Every number is included in those sums 3 times, so s1+s2+... + s10 = 3 x (a1 + a2 +...+ a10) = 3 x 5 x 11=165

165<170 but it fits, there is no contradiction of my assumption. Can I conclude that there is at least one distribution of the 10 numbers where there is not even one string of three consecutive numbers whose sum is at least 18?

Did I prove the contrary of what was asked to be proven?

RuWel