TITLE: A wheel has the numbers 1 to 25 randomly placed on it. Show that there are three adjacent numbers whose sum is at least 39.
QUESTION [5 upvotes]: Any thoughts on understanding how to do this using the Principle of Mathematical Induction would be great.
A wheel of fortune has the integers from 1 to 25 placed on it in a random manner. Show that regardless of how the numbers are positioned on the wheel, there are three adjacent numbers whose sum is at least 39?

REPLY [3 votes]: Note that this can be improved to 41 (instead of 39) using the same ideas.
Consider the position of the number 1.
Take the 8 consecutive sets of 3 digits after that. (These sums would correspond to $S_2, S_5, S_8, S_{11}, S_{14}, S_{17},S_{20}, S_{23}$  in MJD's notation.)
These sets have a sum of $ 2 + 3 + \ldots + 25 = 324$.
Hence, by the pigeonhole principle, one of them must have sum at least $\left\lceil \frac{324}{8} \right\rceil = 41 $.
Is 41 the best we can do? No idea.