Find all series of consecutive positive integers whose sum is exactly 10,000.
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What if we don't require the consecutive integers to (all) be positive?

In general, for there to be a series of consecutive integers with n terms that adds up to some number x, two things are true:

For odd n, x must be a multiple of n. Thus, the terms in the sequence are centered around x/n, with (n - 1)/2 terms on either side of x/n.

For even n, (x - n/2) must be a multiple of n. For this to be true, x must be a multiple of n/2, but not a direct multiple of n. In that case, the terms in the sequence are again centered around x/n (which ends in .5), with n/2 terms above and n/2 terms below.

For the specific case of 10000, we could just start at one and test each number to see if it works, but we can make things a little easier.

First of all, the prime factorization of 10000 is just 2⁴ × 5⁴, so the only odd factors will be powers of 5 (and, of course, 1).

For n = 1, you have of course a series of one term in 10000 itself.

10000 = 10000

For n = 5, there is a series of 5 terms centered around 10000/5 = 2000:

1998 + 1999 + 2000 + 2001 + 2002 = 10000

For n = 25, you have a series of 25 terms centered around 10000/25 = 400:

388 + 389 + 390 + 391 + ... + 409 + 410 + 411 + 412 = 10000

For n = 125, there is a series of 125 terms centered around 10000/125 = 80:

18 + 19 + 20 + 21 + ... + 139 + 140 + 141 + 142 = 10000

For n = 625, there is a series of 625 terms centered around 10000/625 = 16. Note that this series only counts when we take away the restriction that all terms of the series must be positive.

-296 + -295 + -294 + -293 + ... + 325 + 326 + 327 + 328 = 1000

For even n, we need to find values for which n/2 is a factor of 10000, but n itself is not. Based on the prime factorization of 10000 (= 2⁴ × 5⁴), our values need to include all four factors of two, plus an extra 2 to make n itself indivisible into 10000; and any number of factors of five (between zero and four). So, we have five possibilities: 32, 160, 800, 4000, and 20000.

For n = 32, there is a series of 32 terms centered around 10000/32 = 312.5:

297 + 298 + 299 + 300 + ... + 325 + 326 + 327 + 328 = 10000

For n = 160, there is a series of 160 terms centered around 10000/160 = 62.5:

-17 + -16 + -15 + -14 + ... + 139 + 140 + 141 + 142 = 10000

For n = 800, there is a series of 800 terms centered around 10000/800 = 12.5:

-387 + -386 + -385 + -384 + ... + 409 + 410 + 411 + 412 = 10000

For n = 4000, there is a series of 4000 terms centered around 10000/4000 = 2.5:

-1997 + -1996 + -1995 + -1994 + ... + 1999 + 2000 + 2001 + 2002 = 10000

Finally, for n = 20000, there is a series of 20000 terms centered around 10000/20000 = .5:

-9999 + -9998 + -9997 + -9996 + ... + 9997 + 9998 + 9999 +10000 = 10000

So, here is the complete list, with those series containing only positive integers listed first:

10000 = 10000
1998 + 1999 + ... + 2001 + 2002 = 10000
388 + 389 + ... + 411 + 412 = 10000
297 + 298 + ... 327 + 328 = 10000
18 + 19 + ... + 141 + 142 = 10000

-9999 + -9998 + ... + 9999 +10000 = 10000
-1997 + -1996 + ... + 2001 + 2002 = 10000
-387 + -386 + ... + 411 + 412 = 10000
-296 + -295 + ... + 327 + 328 = 1000
-17 + -16 + ... + 141 + 142 = 10000

Looking at these series, a correlation between the positive and negative series should be obvious. Each of the negative series has a number of terms that simply add up to zero and cancel each other out, leaving us with one of the positive series that we have already found.

Posted by DJ on 2004-07-21 13:25:35