In the last column, we looked at the puzzle of determining if the sum of N consecutive integers is divisible by N. This puzzle and one way of solving it is given here.

Based on some quick attempts to verify the conjecture (i.e. the sum is evenly divisible by N), we saw conflicting evidence when we used runs of four or five integers. We might have been a bit confused when allowing negative integers and found that we always get a sum of zero for N odd if the sequence is symmetrically placed about zero. But that was a red herring because zero is okay for the sum since it can be divided by N with no remainder.

Much of this series of articles is about ways and methods to solve problems and puzzles. So I encourage readers to visit the above site to see how one person solved this puzzle - or, more accurately, how one person chose to present the answer since we do not know the process used to get there. And that is the problem with many puzzles. We are given the puzzle and maybe an answer, but often the way of deriving the answer that was actually used has nothing in common with the cleaned version for presentation. Puzzle workers can get the impression that their messy solution - obtaining methods are sub-standard because they never follow the concise, logical steps that an author presents.

Since my way of solving this puzzle leads to the same answer as given by Jim, we can expect that some of the development is the same. However, I did do it a bit differently.

To start, as with the EVE puzzle, I tried to simplify the puzzle by changing it into something that looks simpler. For a string of N integers, subtract the first integer from each of the following ones. Then you have N identical integers plus some left over stuff. Since N times any sum is divisible by N, we need only look at the leftovers.

By the conditions of the puzzle, the left overs form an arithmetic series of N terms 0, 1, 2, 3… (N-1). But we know the formula (or can easily derive it) for that sum. It is simply (N)(N-1)/2. So if we divide this sum by N, we get (N-1)/2. From this we see that if N is odd, the ratio is an integer. If N is even, the ratio is a fraction. Therefore if N is even, the sum on N consecutive integers is not evenly divisible by N, but if N is odd, it is.

Next, I suggested that we consider extending the basic puzzle. For instance, why limit it to integers? After a few moments of thought, we show that the puzzle works if we consider summing a sequence of real numbers that differ by one. That is, the we need not confine ourselves to integers, but the key is the difference between the terms.

The next extension one might logically do is to consider cases where the difference between adjacent terms is not limited to one, but can be any integer.

Review the two ways to solve the original puzzle and see if either way can be extended easily to the more general case. We’ll go through it in a later posting.

What other extensions can be made? Would relaxing the condition of having integer differences make an interesting puzzle? Is anyone ambitious enough to generalize this puzzle to complex numbers?

Sometimes constructing a meaningful puzzle is as much of a challenge as solving puzzles.

For those who wish to delve further into decision theory without wading through a lot of equations, I have posted a tutorial on elementary decision theory. It shows examples of faulty physicians’ diagnoses (important for those considering surgery) and how to evaluate anti-terrorist activities (important for everyone). That tutorial can be found here.