TITLE: Numbering students inequality problem
QUESTION [6 upvotes]: Ten students are sitting around a campfire. A teacher randomly assigns each student a different number from 1-10. Another teacher assigns a new number to each student with the requirement that the new number assigned to a student is equal to the students' old number plus the sum of the two numbers of his neighbors. Prove or disprove: one of the students must have a new number STRICTLY GREATER than 17. 
What I have done so far: It is very easy to show that one of the students must have a new number strictly greater than 16. This can be done by assuming that each of the numbers is less than or equal to 16. A quick sketch is as follows: 
Label the students $a_1, \dots a_{10}$. By contradiction, we have:
$a_1+a_2+a_3 \le 16$
$\dots$
$a_{10} +a_1+a_2 \le 16$
Summing up:
$3a_1+\dots +3a_{10} \le 16\cdot 10$
This is a contradiction as $a_1+\dots + a_{10} = 55$
But this argument breaks down when we try it for 17. Im not even sure that the proposition that a student must have a number more than 17 is even correct! Can someone prove it or find a counterexample?

REPLY [2 votes]: Using ideas from Arturo Magidin's answer and Mark Bennet's comment:
The average score in the first round is $5.5$, so the average score in the second round is $16.5$.  Since scores are integers, this means at least one of the following is true:

Someone has strictly more that 17 in the second round 
At least two neighbours have exactly 17 in the second round
The arrangement must have alternating 17s and 16s

For (2), look at $a_1+a_2+a_3 = 17$ and $a_2+a_3+a_4 = 17$, to solve for $a_4$ in terms of $a_1$.
For (3), look at $a_1+a_2+a_3 = 17$ and $a_2+a_3+a_4 = 16$,  (ignore  $a_3+a_4+a_5 = 17$,)  and at $a_4+a_5+a_6 = 16$ and $a_5+a_6+a_7 = 17$, to solve for $a_4$ in terms of $a_1$ and then to solve for $a_7$ in terms of $a_4$.
So (1) must be the case.