Two little puzzles for today:

1) Find nine conscutive integers whose sum is 6.

2) How many ways can 1000 be written as the sum of consecutive integers?

January 30, 2007 pm31 8:58 pm 8:58 pm

I must be missing something…

If n is the first of the nine consecutive integers, then the sum of those nine integers is 9n+36, and neither 9n+36=6 nor 9n+36=1000 have integer solutions for n.

2. January 30, 2007 pm31 10:17 pm 10:17 pm

I should have clearly separated the two problems.

The second problem doesn’t specify how many consecutive integers. For example:
7+8 = 15, but also 4+5+6 = 15

3. January 31, 2007 am31 12:06 am 12:06 am

For the first question:

n + (n + 1) + (n + 2) + (n + 3) + (n + 4) + (n + 5) + (n + 6) + (n + 7) + (n + 8) = 9n +36 = 6.

That doesn’t have any integral solutions. I’ll have to check in later to see whether I miscalculated or missed a gimmick.

On the second one, there’s 198, 199, 200, 201, 202. There’s one from 28 to 52, one from -54 to 70. And I’ll try to find time to generalize it and check for other solutions. That’s the hard part, of course.

My late lunch is over, back to work.

January 31, 2007 am31 12:48 am 12:48 am

Problem 1, as rdt says, is a wild goose chase.

Problem 2 is great.
For any number N, I believe the answer is twice the number of distinct odd factors of N (including 1).

January 31, 2007 am31 3:24 am 3:24 am

So, doing the correct version of the second problem… The sum of k consecutive integers starting at n is k(n-1)+k(k+1)/2.

So the problem is to find the pairs (n,k) which satisfy 1000= k(n-1)+k(k+1)/2. John’s (198,5), (28,25) and (-54, 125) all work…

The expression for k turns into the quadratic k^2+k(2n-1)-2000=0.

More later… This is much too good a way to avoid tidying the house…

6. January 31, 2007 am31 3:24 am 3:24 am

#1 can be nice for kids to demonstrate (w/o algebra) that there is no integer solution
-3 + -2 + -1 + 0 + 1 + 2 + 3 + 4 + 5 = 9, too big
-4 + -3 + -2 + -1 + 0 + 1 + 2 + 3 + 4 = 0, too small
This was not going to be a stumper for you guys.

#2 is trickier, I think.

January 31, 2007 am31 3:59 am 3:59 am

Picking up my train of thought, though only briefly…

The quadratic K^2+k(2n-1)-2000 has integer solutions only if 4n^2-4n+8001 is the square of an odd number (for n=198, it’s the square of 405).

Which means 4n^2-4n+8001 = (2j+1)^2… I’m thinking that’s either going to have 4 integer solutions, or infinitely many…

January 31, 2007 am31 6:06 am 6:06 am

Oops, no, I didn’t properly absorb Clueless’s comment — it’s twice the number of odd factors…

4n^2-4n+8001 = (2j+1)^2 can be re-written (n+j)(1+j-n)=2000, which, substituting r=n+j can be written r(1+r-2n)=2000. Or for the general case, r(1+r-2n)=2C

This has integer solutions for r,n if, and only if r is odd. All the odd factors of 2C are odd factors of C, and r can be either positive or negative.

In the case of 1000=2^3 x 5^3, the odd factors are 5, 25 and 125, so there should be 6 solutions.

January 31, 2007 am31 6:07 am 6:07 am

Ack, no, the odd factors of 1000 are 1, 5, 25 and 125 so there should be 8 solutions.

10. January 31, 2007 am31 7:04 am 7:04 am

So there are 8 solutions for 15?

(first integer,number of integers)

[15,1]
[7,2]
[4,3]
[1,5]
[0,6]
[-3,10]
[-6,15]
[14,30]

OK, that works. When I’ve run this problem, we’ve generalized a solution for powers of 10 only. I like your (both of you) more general approach.