SOLUTION: Express 1994 as a sum of coonsecutive positive integers and show this is the only way to do it.

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Question 42055: Express 1994 as a sum of coonsecutive positive integers and show this is the only way to do it.
Answer by kev82(151) About Me  (Show Source):
You can put this solution on YOUR website!
Hi,
1994 can be written as 497+498+499+500 so there is how it's done. Proving this is the only way is quite simple but it's quite long and I don't really want to type it here. An outline of the proof is as follows.
1) To write a number N as the sum of (possibly negative) consecutive integers, where the first term is a. Then
N=na%2Bn%28n-1%29%2F2
2) We can rearrage to find a
a=%282N-n%28n-1%29%29%2F2n
3) We want a to be an integer so (replace equals with congruent to)
n%28n-1%29=2N+mod+2n
4) (Be careful with 2N congruent 0 mod 2n here!) If n is odd then this has a solution when n divides N. If n is even then there is solution when n does not divide N, but does divide 2N.
5) So taking N=1994, the only values of n that satisfy the above conditions are n=4,997,3988. (N=1 is obviously a trivial solution, but you really want n>1 right?)
6) Calculating a for the above 3 values gives 497, -496, and -1993. You are only interested in the sums which use strictly posotive numbers. There is only one of those, with n=4 which I gave you above.
Hope that helps,
Kev