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The problem is: Prove that any circular permutation of 1, 2, ..., 36 has a sum of five consecutive numbers greater than or equal to 95.

Since the average sum of any five consecutive numbers is (1+2+...+36)×5/2=92.5, I can prove that the answer is greater than or equal to 93. But I can't reach 95, nor can I provide an example where the maximum sum is exactly 95. Can someone help me with this problem?

log2
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    You might try searching for similar problems on MSE. For example, this may be of interest: https://math.stackexchange.com/questions/507920/ – Kyle Apr 06 '24 at 22:40
  • @Kyle Thank you! That really helps. – log2 Apr 06 '24 at 23:20

1 Answers1

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I assume that 'circular permutation' means placing the numbers on a circle in an arbitrary order? In particular, if you were to label the numbers as $a_1, \ldots, a_{36}$, we're allowed to consider the sum $a_{35} + a_{36} + a_{1} + a_{2} + a_{3}$, right? I will work with this assumption.

You have the correct idea. You want to take an average and argue that the maximum is at least as large as the average. The trick that I present here only works because of the specific numbers in question. We will pick a special set to average over, so as to get a larger average.

Ignore the number 1 on the circle, and you are left with 35 consecutive numbers. Group them into 7 consecutive intervals of 5. These have a total sum of $\frac{36 \times 37}{2} - 1 = 665$, and thus have an average value of $\frac{665}{7} = 95$. There you go!

This bounding was so convenient because $5$ is a factor of $36-1$. Otherwise, you don't have much control over the integers you wish to discard. For example, if we were working with $1, \ldots, 37$, and I discard two numbers, I can choose for one of the numbers to be one, but I cannot choose the other number now. It might as well be $37$, in which case, our method doesn't even give a sharper upper bound (verify this!).

yeetcode
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  • Thank you! It's a nice answer! Further, can we reach 96 or more in my question? – log2 Apr 06 '24 at 23:11
  • The 1,2, ... , 37 problem is interesting. In this case, can we prove 95? – log2 Apr 06 '24 at 23:17
  • Well, you can prove that the answer cannot decrease. Take any arrangement for 1-36. There are 5 consecutive entries adding up to at least 95. Check that adding 37 in the middle allows you to get a strictly larger sum. On the other hand, not adding 37 in the middle leaves this tuple's sum unchanged.

    I am not sure about being able to reach 96 in your question. Maybe it's possible, but I don't know.

     – yeetcode Apr 07 '24 at 10:57