The integer 30 can be written as a sum of consecutive positive integers in three ways:
30 = 9+10+11 = 6+7+8+9 = 4+5+6+7+8.

Find the smallest positive integer which can be written as a sum of consecutive positive integers in 12 ways.

I found a lower number with more sequences, 1575.  It has 17 sequences with the following lengths:

2,3,5,6,7,9,10,14,15,18,21,25,30,35,42,45,50

Those sequences are:

1575=
787-788
524-526
313-317
260-265
222-228
171-179
153-162
106-119
98-112
79-96
65-85
51-75
38-67
28-62
17-58
13-57
7-56

I plan to search a little further.

Now to ramble about my method (which I thought out as I typed).

Notice that if you average a sequence of consecutive integers, you will always get a multiple of 1/2, a whole number when the length of the sequence is odd, and a fraction when the length is even.

So, if an integer N can be written as the sum of a sequence of consecutive integers L terms long, then N/L is a multiple of 1/2.  Also, it will be a whole number iff L is odd.

Finally, notice that given the above, the smallest possible N is L(L+1)/2 (the triangle number of L).  The inverse of this function is "L is less than or equal to (1+sqrt(1+8N))/2."

So the number we're looking for is half of the smallest integer N with 12 divisors whose triangle numbers are smaller than N/2.  These divisors cannot include numbers that include some but not all of the 2 factors in N, or the number 1.

I don't know how to minimize such a number, but I do know that the number of divisors is equal to the product of one plus each of the exponents in the number's prime factorization.  This gave me a starting point, and the above ramblings gave me a good way to test it.

Edited on June 28, 2004, 6:13 pm

Edited on June 28, 2004, 6:29 pm

Edited on June 28, 2004, 6:30 pm

Posted by Tristan on 2004-06-28 18:13:13