Consecutive Integers Calculator

One equation, not a search

The whole method is naming the first term. Once n is fixed, every other term is fixed with it, and a problem that looked like a search becomes one linear equation.

Three consecutive integers are n, n+1 and n+2. Their sum is 3n + 3, so 3n + 3 = 45 gives n = 14. No trial and error, and no guessing.

In general, k terms starting at n sum to kn + k(k−1)/2 — the k copies of n, plus the offsets 0 through k−1, which add to the triangular number k(k−1)/2.

Because the equation is linear, there is at most one run of a given length. It either exists or it does not, and there is never a choice between two.

When there is no answer

Three consecutive integers summing to 100 do not exist, and the reason is worth more than the answer.

With an odd count the middle term is the mean of the run. So the sum must be exactly the count times a whole number — a multiple of the count. 100 is not a multiple of 3, so there is nothing to find.

With an even count there is no middle term; the mean falls between the two central values and lands on a half. So the sum is the count times something-and-a-half, which makes it an odd multiple of half the count. That is why eight consecutive integers CAN sum to 100 — they are 9 through 16.

A page that answered “no solution” and stopped would be hiding the useful part. The condition tells you which run lengths to try before you try any of them.

Even and odd runs

“Three consecutive even integers” is a different problem, and it needs two conditions rather than one.

The step is 2 rather than 1, so k terms from n sum to kn + k(k−1). Solve that for n as before.

But n also has to be the right kind of number. Three consecutive odd integers summing to 60 fails not because the division does not work — it gives 18 — but because 18 is even, and 18, 20, 22 are not odd numbers. The arithmetic succeeds and the parity check does not.

There is a shortcut for odd counts here too: three consecutive odds sum to three times the middle one, and the middle one has to be odd. 57 ÷ 3 = 19, which is odd, so 17, 19, 21 works.

The powers of two

Here is a result that turns an exercise into something worth knowing.

Every positive whole number can be written as a sum of two or more consecutive positive integers — except the powers of two. 1, 2, 4, 8, 16, 32 and so on cannot be, and everything else can.

15 = 7+8 = 4+5+6 = 1+2+3+4+5, three different ways. 64 has none at all.

The reason is that a run of k terms starting at n sums to k(2n + k − 1)/2, and of the two factors k and (2n + k − 1) exactly one is always odd. So any number with a representation has an odd factor above one — and the powers of two have none.

The number of representations is exactly the number of odd divisors above one, which is why 15 — with odd divisors 3, 5 and 15 — has three.

Where it gets used

Word problems. The standard vehicle for teaching “define a variable, then express everything else in terms of it”, which is the single most transferable idea in school algebra.

Puzzles. Page numbers on a folded sheet, houses on one side of a street, seats in a row — all consecutive-run problems in disguise.

Number theory. The powers-of-two result is a genuinely nice theorem reachable from a school exercise, and the counting version connects it to divisors.

Triangular numbers. A run starting at 1 IS a triangular number, and the k(k−1)/2 in the formula is where they come from.

Sources and methodology

The sum formula and the powers-of-two result are standard; these are the references.

• American Mathematical Society — where sums of consecutive integers and the powers-of-two result are catalogued
• National Council of Teachers of Mathematics — where these word problems sit in the curriculum
• Arithmetic Sequence Calculator — the general case, of which this is the step-1 version
• How we calculate — methodology

Method. The first term is solved for directly rather than by searching, and the divisibility test that decides whether a run exists is done on integers so a near-miss cannot round into a false answer. The suite does not trust the closed form: it re-derives every case from 1 to 300 across run lengths 2 to 8 by exhaustive search and compares, and separately checks that the terms the page actually prints sum back to the target. That engine is verified on every change against 155 hand-written assertions, including that the closed form agrees with an exhaustive search across 2,100 cases, and that only the powers of two admit no run at all, checked to 512. The count and the per-case breakdown are published on the formula verification page.

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Read the guide

A consecutive run is an arithmetic sequence with a common difference of one, which is why its sum has a closed form: n terms starting at a add to n·a + n(n − 1)/2. That also settles which totals are reachable at all — a sum of three consecutive integers is always a multiple of three, and a sum of an even count of them never is, which is why some of these problems have no solution rather than a fractional one.