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. 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?
 A: I don't know how you'd apply induction to this, but it's not a hard problem. 
There are 25 3-number segments. (They overlap one another, of course.)  Let $S_1, S_2, \ldots, S_{25}$ be the sum of the three numbers in each segment.  If you add up $S = S_1+\cdots + S_{25}$ you have added each of the numbers $1,\ldots, 25$ three times, so you can calculate exactly what $S$ must be.
Now suppose each of $S_1,\ldots, S_{25}$ were less than 39.  This would put an upper bound on how big $ S = S_1+\cdots + S_{25}$ could be.  Would this be consistent with the value of $S$ you found in the previous paragraph? If not, you've showed there must be some $S_i$ that is at least 39.
A: 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.
