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?

The sum of an arithmetic series is n*[2a+(n-1)*d]/2. Since the numbers are consecutive, d=1.

n*(2a+n-1)/2=10000
n*(2a+n-1)=20000

a=[(20000/n)-(n+1)]/2

Since we are dealing with integers, n must be a factor of 20000 and by definition is positive. Also, a must be an integer too (I'll do both parts at the same time).

20000 can be factored into (2^5)*(5^4). If n were odd, then a would automatically be an integer. If n were even, then the only way a can be an integer is if 20000/n is odd. From the factors of 20000, the only possibility is if the factor is (2^5)*(5^k), where k is 0, 1, 2, 3, and 4.

There are 5 odd factors for 20000 and 5 possibilities for (2^5)*(5^k), giving a total of 10 series.

n=1, a=10000
n=5, a=1998
n=25, a=388
n=125, a=18
n=625, a=-296
n=32, a=297
n=160, a=-17
n=800, a=-387
n=4000, a=-1997
n=20000, a=-9999

Organizing it, that gives

POSITIVES

{10000}
{1998, 1999, 2000, 2001, 2002}
{388, 389, ... , 411, 412}
{18, 19, ... , 141, 142}
{297, 298, ... , 327, 328}

NEGATIVES

{-296, -295, ... , 327, 328}
{-17, -16, ... , 141, 142}
{-387, -386, ... , 411, 412}
{-1997, -1996, ... , 2001, 2002}
{-9999, -9998, ... , 9999, 10000}

Posted by np_rt on 2004-07-21 16:41:48