Comments on: Consecutive Integers Puzzle from JD2178

By: Jonathan

Jonathan — Sun, 08 Nov 2009 12:33:56 +0000

It could be interpreted either way… so let’s choose the way that makes things work nicer?

By: Nick

Nick — Sun, 01 Nov 2009 20:28:28 +0000

I was a little confused as to whether or not 1000 by itself could constitute a “sum of consecutive integers” in the phrasing of the problem, thoughts?

By: watchmath

watchmath — Fri, 23 Oct 2009 22:13:22 +0000

I can prove that there are 8 solutions. I believe you missed the sequence that consist of single number which is 1000 itself. Let me know if you want to see the prove (the prove is simple).

By: Nick

Nick — Mon, 19 Oct 2009 05:18:08 +0000

Hi Jonathon, sorry ’bout that, it’s now fixed.

By: Sum of Consecutive Integers | Mathing...

Sum of Consecutive Integers | Mathing... — Sun, 18 Oct 2009 06:47:12 +0000

[...] first came across the first puzzle through Nick’s blog Divide By Zero where he found the form of the solutions to the puzzle. He [...]

By: Jonathan

Jonathan — Sun, 18 Oct 2009 00:41:16 +0000

Hi, that's jd2718 (as in e), not 217. I posted an alternate solution. And then I posted a tougher follow-up question that you might enjoy! Jonathan

By: Puzzle Exension – Consecutive Integers « JD2718

Puzzle Exension – Consecutive Integers « JD2718 — Sun, 18 Oct 2009 00:40:31 +0000

[...] “How many ways can 1000 be written as the sum of consecutive integers?“ Nice puzzle posted here a few days ago. We had some nice discussion and partial and complete solutions in the comments. Nick linked back to a longer discussion on his blog, Divide by Zero. [...]

By: Alex

Alex — Wed, 14 Oct 2009 19:15:59 +0000

One final note (sorry to take up 3 posts! Feel free to smoosh them together if that’s possible) – I probably won’t look back here, so if you find mistakes or have a question, I’m on gmail as alexrdavies1

By: Alex

Alex — Wed, 14 Oct 2009 19:11:52 +0000

Before, we said 5+6+7+8+9 = 7*5, so 5 divides the answer.
Now, we have 5+6+7+8 = 6.5*4 = 26. We clearly can’t conclude that 4 divides 26 – but is true that 4/2 has to divide 26.
Why is that so? Well,

(sum of an even number of integers) = midpoint * length = 2*midpoint * length/2 = sum.
Thus half the length (remember the length is even here) divides the sum. But there’s more: crucially, the midpoint can’t be a whole number, so 2*midpoint MUST BE ODD. Once again we have a 1:1 correspondence between odd factors, and sequences.

Again, the odd factors of 1000 are 1, 5, 25 and 125. These correspond to lengths of
(1000/1)*2 = 2000, giving a sequence …..+0+1+…..
(1000/5)*2 = 400, giving a sequence …..+2+3+…..
(1000/25)*2 = 80, giving a sequence …..+12+13+…..
(1000/125)*2 = 16, giving a sequence …..+62+63+…..

So in total, there eight sequence of integers summing to 1000 (though I did count “1000″ as an answer, you might not).
More generally, though, for any number, there will be twice as many of these sequences as the number has odd factors.

So for example, 1,000,000 = 2^6 * 5^6 has seven odd factors, so it can be expressed as a sum of consecutive integers in 14 different ways.

By: Alex

Alex — Wed, 14 Oct 2009 18:59:32 +0000

Here’s my attempt at a different solution. The first part came very naturally:

First, note that 5+6+7+8+9 = 7*5. Why does this matter? Well, in general the sum of (2k+1) consecutive integers will always be a multiple of (2k+1). Indeed, there will be at ONE SUCH SUM for EVERY ODD FACTOR of our target number.

So, what are the odd factors of 1000? Looking at its prime factorisation, 2cubed * 5cubed, we see quickly that they are 1, 5, 25 and 125.

Thus all the sums of odd length are
1: 1000
5: 198+199+200+201+202
25: …..+40+…..
125: …..+8+….. – note that a lot of these integers are negative.

So much for sums of an odd number of integers. Sums of an even number of integers are a little more complicated, though.