Expert: Sherry Wallin - 6/15/2011

Question
what is the largest possible value of n (n is an odd number) if n consecutive positive numbers are added to get a sum of 2007th

Answer
Hi Leonard~


I had to read this problem more carefully. You need to be clearer. Is it n consecutive positive integers, and n can be either even or odd, OR is it that n is odd and you want n consecutive ODD INTEGERS?

In the first scenario, restricting n to be odd makes there be an odd number of consecutive integers and the list will begin with an odd integer and end with an odd integer. But this will give you an EVEN sum and thus you will never get 2007. Try it with a few lists of consecutive integers beginning with an odd integer and thus ending with an odd integer. Or take my word for it that an
even + odd = odd and an odd + odd = even. So an odd + even + odd = (odd + even) + odd = odd + odd = even.  By the way, what do you mean by 2007"th"?

Hence, I have come to the conclusion that it is the 2nd scenario but you haven't asked the question that way! If we have an odd number of integers that are consecutively odd then you will have odd + odd = even and (odd + odd) + odd = even + odd = odd and so on and so forth...


First let's factor 2007 = 3^2 * 223
You can see from the factorization that there are only 2 prime factors of 2007, that is 3 and 223. You have as 'other' divisors 1, 3, 3*3 = 9, 3*223 = 669, 2007. We can eliminate n = 2007 since that is the number we are trying to get the sum for and it is impossible to add 2,007 consecutive positive integers to get to a sum of 2007. Likewise we can omit 1, because 1+3+5+...+4013+4015 far exceeds 2007. (Notice there needs to be 2*2007 + 1 in the list to be summed if we are adding 2007 consecutive odd integers).. Again we can omit 669 because 3*669 = 2007 so there is no way to have 669 consecutive odd integers to sum to  2007. I think you would agree that even 223 is not going to happen because  9*223= 2007 and 223^2 = 49729 (if we were counting consecutive integers and not consecutive odd integers there would be all the integers from 223 and then the next 49,729 integers to be summed and since we are talking about consecutive odd integers that means we would be summing twice that many or 99458 + 1 odd integers beginning with 223 which far exceeds the sum we are looking for). This actually just leaves 3 and 9 consecutive odd integers and since we are looking for the largest let's just see what we can find if n = 9:

n + (n + 2) + (n + 4) + (n + 6) + (n +8) + (n + 10) + (n + 12) + (n + 14) + (n + 16) = 9n + 72 which has to equal 2007, thus solving for n:
9n + 72 = 2007
9n = 1935
n = 215
Try it out: 215 + 217 + 219 + 221 + 223 + 225 + 227 + 229 + 231 = 2007
Therefore, 9 consecutive odd integers is the largest odd number of consecutive odd integers that sums to 2007

Math Prof