# Maximum sum of 3 consecutive numbers in a permutation [closed]

Given that $X = \{0, 1, 2, ..., 7, 8, 9\}$, and $P$ is a permutation on $X$. Let $M(P)$ be the maximum sum of 3 consecutive elements. For example, if $P = (0, 2, 4, 1, 5, 7, 9, 3, 8, 6)$, then $M(P)$ is the maximum integer among $6, 7, 10, 13, 21, 19, 20, 17$ (which are the sums of each 3 consecutive elements, respectively). In the example above, $M(P) = 21$. Prove that there exists no permutation $P$ that satisfies $M(P) = 12$. Also, find a way to construct a permutation that satisfies $M(P)=13$. Any help would be extremely appreciated :)

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## closed as too localized by Mark Sapir, Bill Johnson, Douglas Zare, Igor Rivin, Andreas ThomJan 7 '12 at 20:44

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I am sorry, but this is certainly not a research question, and hence (as far as I understand the purpose of this forum) not a suitable question for mathoverflow. I suppose this thread will be closed within the next few minutes (and rightfully so).

However, since I read the question and started thinking about it, here is a solution to your problem (which, I hope, is not your homework):

You have $M(P) = 13$ for $(9,4,0,8,3,2,7,1,5,6)$ and it is not hard to prove that you have $M(P) > 12$ for all $P$: Assume that we have $M(P) \leq 12$ for some $P = (n_0,\ldots,n_9)$. Then,

$\underbrace{n_1 + n_2 + n_3}_{\leq 12} + \underbrace{n_4 + n_5 + n_6}_{\leq 12} + \underbrace{n_7 + n_8 + n_9}_{\leq 12} \leq 36$,

and

$\underbrace{n_0 + n_1 + n_2}_{\leq 12} + \underbrace{n_3 + n_4 + n_5}_{\leq 12} + \underbrace{n_6 + n_7 + n_8}_{\leq 12} \leq 36$.

However, since $P$ is a permutation, we must have

$\sum_{i=0}^9 n_i = 0 + 1 + \ldots + 9 = 45$.

So the two inequalities above imply $n_0 = 9$ as well as $n_9 = 9$, a contradiction.

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Thanks a lot :) I'm sorry if the question isn't suitable, I'm still a newbie (which is quite deducible from my reputation :D). Actually the problem is from a past local Math Olympiad. Anyhow, thanks again for the answer! – somethingsomething Jan 7 '12 at 20:13