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?
Best Answer
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.