tuple of integers

Question

The integers 1,2,...,30 are invited to a dinner party. They all sit around a round table, in some unknown order. Does there exist an ordering in which there are no three successive (successive means sitting next to one another) integers whose sum is at least 45. If there exist such ordering prove it by constructing it. If there does not exist such ordering, prove that for every possible ordering there exist at least one tuple of three successive integers that sum to 45 or more.

Came across this interesting question in a text book and the solution isn't available.

Answer 1

Consider all of the successive sums of triples of integers. There are $30$ such sums.

Since each of the integers appears in exactly $3$ of these sums, the total of all of the $30$ sums is

$$3\sum_{k=1}^{30} k = 3 \cdot 15 \cdot 31 = 45 \cdot 31$$

Note that this sum is greater than $45 \cdot 30$, so at least one of the $30$ sums must be greater than $45$, by the pigeonhole principle.

Answer 2

First, we will find the total sum of all the numbers that are used. $((1+30)/2)*30$. We multiply this by three, since each of the numbers are used three times, when they are first in the set of three, when they are the middle, and when they are at the end. We get 1395. Divide 1395 by the total number of these three consecutive number sets, which is 30. 1395/30 is 46.5. This 46.5 is the average number of the sum of the consecutive sets. It is greater than 45, which shows that at least one set must have a sum greater than 45.