When a sum of consecutive numbers is a multiple of something

0:00Here are six sums. Each is a run of numbers that follow straight on from each other. Seven is three plus four. Ten is one plus two plus three plus four. Twelve is three plus four plus five. And fifteen turns up three times: seven plus eight, four plus five plus six, and one plus two plus three plus four plus five. Six different runs, and one number done three ways while the others are done once.

0:31That is enough to start asking. Can every counting number be written like this? Which ones more than one way? Every odd number is a pair of neighbours, so what about the even ones? And if negatives are allowed in, is zero one of these sums? By the end, all four. Start with the smallest run there is. Two numbers side by side. Three and four make seven. Eight and nine make seventeen. Twenty and twenty-one make forty-one.

1:04Every one of those totals is odd, and that is not luck. Write the pair as some number and the next one up. Together they come to twice that number, plus one. Twice anything is even, and one more than even is odd. So a pair of neighbours is odd, always. Checked across sixty starting points, and again over a range reaching down past zero. Never once even. Which settles one of the questions. No even number is a sum of two consecutive numbers; if it is one of these sums at all, it needs three terms.

1:44So try three. Four, five and six. They come to fifteen. Look at the outer two. Four is one below the middle, six is one above. The short one and the over one cancel exactly, and the run levels off into three fives. Three times five is fifteen. The total is three middles. That works for any three in a row: the outer two are always one below and one above.

2:13So the total of three consecutive numbers cannot help being a multiple of three. Sixty starting points, and sixty multiples of three. Now five, and nothing changes except how far the levelling reaches. One, two, three, four, five. Fifteen again, by a different route. The middle is three. Two is one below and four one above; one is two below and five two above. Both pairs cancel, and the run levels off into five threes. Five times three is fifteen.

2:50Any odd length does this. There is a genuine middle term, the rest pair off around it, and every pair cancels. So an odd-length run totals its length times its middle, and its total is a multiple of its length. Checked on a hundred and twenty runs, and then on two hundred and sixty-five more. Now four, and something breaks. Three, four, five and six. The total is eighteen. Find the middle term. There isn't one.

3:24Four terms have a gap in the middle, not a term. The average sits at four and a half, which is not in the run. So there is nothing for the total to be four of. And it is not. Eighteen is even, but eighteen is no multiple of four. Nor is any other four-term total. Eighty starting points, not one a multiple of four. The count of terms is simply not a factor.

3:53Which raises the real question of this whole topic. How many runs of four did we just look at? Eighty. How many are there? No end of them. Checking is not the same as arguing, and here is a pattern that shows why. Every number up to eight has at most one way of being written like this. Some of them have no way at all. It looks settled, the sort of thing you write down as a rule.

4:21Then nine arrives, as four plus five and as two plus three plus four. Nine has two. Eight cases in a row proved nothing. To settle something about every run at once, we have to write down every run at once. So here is every run there is, on one line. Call the first term n and the number of terms k. The terms run from n up to n plus k take away one.

4:51Add them. There are k copies of n, so that is k times n. On top of those sit the extras: nothing, then one, then two, up to k take away one. Those add to k times k take away one, halved. For four terms that is six; for five terms, ten. Put the pieces together and tidy up, and the total is k times, two n plus k take away one, all over two.

5:19That is every run, of any length, from any start, written once. Everything left is read off it. Read it first for an odd length. If k is odd then k take away one is even, and two n is even as well. So the bracket is even, and halving it lands exactly on a whole number - checked rather than assumed, and it holds every time. Half of two n plus k take away one is n plus k take away one over two, and that is precisely the middle term.

5:55Which leaves the total as k times the middle. The picture and the algebra are saying the same thing. And the total is a multiple of the length for the same reason every time: the length is sitting there as a factor. Now read the same line for an even length, and watch the factor move. If k is even then k take away one is odd, and two n plus an odd number is odd.

6:24So this time it is the length that gets halved, and the bracket that stays whole. The total is half the length, times something odd. For four terms that is two, times two n plus three. Two times an odd number. Even, because of that two. Never a multiple of four, because the other factor is odd and has no second two to give. That is the missing middle showing up as arithmetic. There was no middle term, so the length never became a factor.

6:59Everything so far has quietly assumed one thing. That each term is one more than the last. Change the step, and watch which results survive it. Take five consecutive even numbers. Four, six, eight, ten, twelve. They total forty. Forty is five times eight, and eight is the middle. The odd-length result survives untouched: the levelling never looked at the step. Now four consecutive even numbers. Two, four, six, eight. They total twenty.

7:34Twenty is a multiple of four. That result has flipped. It flips whenever the step is even, and holds whenever the step is odd. A rule remembered as add six is wrong the moment the numbers stop being consecutive. All of this runs backwards as well as forwards. Suppose you are told only that four consecutive numbers total thirty-four. Take off the extras that piled up. Nothing, one, two and three, which is six. That leaves twenty-eight to share between four terms.

8:12Seven. The run is seven, eight, nine and ten, and adding them back up gives thirty-four. Or the greatest of five consecutive numbers is p. Then the others are p take away one, two, three and four. But if the middle of five consecutive even numbers is five p, the offsets double. Five p take away four and two, and five p plus two and four. Same question, different step, four different answers.

8:46Which leaves the first question of all. Which numbers are these sums at all? Double the total and it splits into two factors of opposite kinds: one even, one odd. So a number needs an odd factor above one before it can be written as a run - and its number of ways is its count of odd factors, less one. Fifteen has odd factors one, three, five and fifteen. Four of them, less one, is three ways. Exactly the three we started with.

9:19The numbers with no odd factor above one are the powers of two. One, two, four, eight, sixteen, thirty-two - none of them is one of these sums. Zero is the honest one. Among counting numbers it is not a sum. Allow negatives and minus one, zero, one is zero, as is minus two through two. The answer depends on the rules. Sums are pinned to their middle. So are products: two in a row make a multiple of two, three in a row a multiple of six, four in a row a multiple of twenty-four.