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.

(In reply to re(2): Computer solution by Charlie)

Sorry, I was wrong in my interpretation of what your program showed. 

3^12 has 13 odd divisors the way they are usually counted (1 is counted, and the number itself is counted if it is odd). The problem statement rules out the trivial one-term representation (consecutive positive integers), so 13 divisors gives 12 non-trivial representations. I guess 3^12 is the smallest number with exactly 13 odd divisors, and if this is so, it is the true  solution (for strict interpreters of 12).

If we allow the trivial representation as one of the 12, then 315 has 12 odd divisors (including 1 and itself) and thus has 12 partitions with consecutive parts.

Edited on June 29, 2004, 3:52 pm

Posted by Richard on 2004-06-29 15:42:34