Prove that the set ${1, 2, . . . , 1989}$ can be expressed as the disjoint union of subsets $A_i (i = 1, 2, . . . , 117)$ such that:
(i) Each $A_i$ contains 17 elements;
(ii) The sum of all the elements in each $A_i$ is the same.
My Solution:
Note that the problem is invariant under translation, so we may centre it at $0$ and it will suffice to show that we can find 117 17 element subsets of the set
$-994,-993,...,994$
with each having its elements sum to $0$.
Now one of the $A_i$ must contain $0$ and hence an even number of non-zero elements, so for this set we can just pair positive with corresponding negative numbers.
The other 116 sets have an odd number of non-zero elements. For each of these, we can take 7 pairs of positive and corresponding negative numbers, but we must then complete the sets with a pair of numbers of a given sign, say $a,b$ and another number of the opposite sign, say $c$, such that $a+b+c=0$.
Now, since we have an even number of sets to construct by choosing elements in this way, we can deal with them in pairs, choosing $a,b,c$ in one, and $-a,-b,-c$ in the other. If we can find 58 such pairs of triplets and assign one triplet to each of the 116 sets, we will then be able to arbitrarily assign 7 positive-negative pairs from the remaining numbers to complete each set.
So the problem thereby reduces to the question of whether we can find 58 disjoint triplets $(p,q,r)$, where $p,q,r$ are distinct positive integers from the set {1,2,...,1994}, such that $p+q=r$. (We would then set $a=p$, $b=q$ and c=-r for a given set).
But this is clearly true, for example,
(100,200,300),(101,201,301),...,(158,258,416)
QED
I'm wondering if this solution is valid and if so would appreciate any suggestions for improvement. Thanks.