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 reached the answer of 945 shortly after my last post.  Here is the logic, briefly:

As Tristan pointed out (but worded differently), a sequence of consecutive integers with an odd number of terms is divisible by that number of terms, while a sequence with an even number of terms is divisible by the number of terms with a remainder of half the number of terms.

Using this, we note that an odd number x that is divisible by 3 can be expressed as the sum of three consecutive integers.  Since x is also divisible by 6 with a remainder of 3, we know it can be expressed as the sum of six consecutive integers.  Thus the solution will most likely be odd, with at least six odd divisors (which implies six more even divisors with a remainder of half the divisor). 

So, if the number is divisible by 3, 5, and 7, it will also be divisible by 15, 21, and 35 -- six odd divisors.  So far our answer is 3*5*7 = 105. 

But these need to be sequences of positive integers, so we need a multiple of 105 that is at least as big as the biggest triangle sum of the set {3,5,7,15,21,35,6,10,14,30,42}. Notice I do not include 70 in this set, since any odd number can be expressed as the sum of two consecutive numbers.

The triangel sum of 42 is 903, so we need 9*105 to meet our condition. Thus one answer is 945.

Since we added mor divisors when we multiplied by 9, we need to check to see if the odd fators from the set of 3*3*3*5*7 yield five distinct odd factors with a maximum less than 21 (remember, we dont need 35*2 to be a factor). Since there is none, 945 is the smallest such number, and our solution.

Posted by Bryan on 2004-06-29 10:11:10