Summing Consecutive Integers

Date: 08/30/98 at 06:05:17
From: Simon
Subject: Arithmetic Sequences

Dear Dr. Maths,

Could you assist me with this problem?

   Express 1994 as a sum of consecutive positive integers and show 
   that this is the only way to do it.

I have worked out the answer by using guess and check methods (the 
answer being 497 + 498 + 499 + 500 = 1994), but I do not know how to 
prove it is the only way. I believe there is something to do with the 
formula for calculating the sum of consecutive numbers in an 
arithmetic sequence.

Thanks,
Simon

Date: 08/31/98 at 12:45:23
From: Doctor Peterson
Subject: Re: Arithmetic Sequences

Hi, Simon. 

You're definitely on the right track, and in the process you've 
observed the big problem with guess-and-check. It's often the fastest 
way to solve a problem, but you don't come away with any confidence 
that you've found the answer.

The formula for the sum of N numbers starting with M; that is, 
M + (M+1) + (M+2) + ... + (M+N-1)  is:

   (2M+N-1)*N
   ----------
       2

as you can easily see by lining up two copies of the sum and adding 
them:

    M       + (M+1)    + (M+2)    + ... + (M+N-1)
   (M+N-1)  + (M+N-2)  + (M+N-3)  + ... +  M
   -----------------------------------------------
   (2M+N-1) + (2M+N-1) + (2M+N-1) + ... + (2M+N-1)

So to make 1994 the sum of a series like this, you have to factor 
2*1994 into two numbers that could be written as N and (2M+N-1); that 
is, into two factors A and B, from which we can get N = A and 
M = (B-A+1)/2.

If I factor 2*1994 = 3988, I get 2*2*997 (997 is prime), so the only 
pairs I can choose to be A and B are

    A      B
   ---    ---
    1     3988
    2     1994
    4      997
    997      4
    1994     2
    3988     1

See what M and N are for each case, and you'll have your proof. (You'll 
notice that if you allow "consecutive positive integers" to be just one 
integer (N = 1) there is another answer!)

I hope this method is clear enough for you. There may be other ways, 
but this is the first way I found.

- Doctor Peterson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/