Question 42055
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+n(n-1)/2}}}

2) We can rearrage to find a

{{{a=(2N-n(n-1))/2n}}}

3) We want a to be an integer so (replace equals with congruent to)

{{{n(n-1)=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