Problem 1

Prove that the set

\{1, 2, \ldots, 1989\}

can be expressed as the union of

117

pairwise disjoint subsets

A_1, A_2, \ldots, A_{117}

, each containing

17

elements, such that the sum of the elements is the same in every

A_i

Step 3 of 5: Build 116 triples that already balance

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In plain words

It's like building $116$ tiny, already-balanced mini-scales out of spare pairs, so later we only need to add matching weights to every scale to keep them level.

\{301,801,1883\}\ \text{and}\ \{1689,1189,107\}\ \text{both sum to}\ 2985=3\times995

Detailed analysis

Take $58$ triples such as $\{301,801,1883\},\{302,802,1881\},\ldots,\{358,858,1769\}$ (first coordinate up by $1$ , second up by $1$ , third down by $2$ each time), together with their $58$ complementary triples $\{1689,1189,107\},\{1688,1188,109\},\ldots,\{1632,1132,221\}$ obtained by replacing each entry $x$ with $1990-x$ . Every triple listed sums to $2985=3\times995$ , because replacing $x$ by $1990-x$ in one coordinate and compensating by $\mp1$ in another keeps the total fixed, and the first triple already sums to $301+801+1883=2985$ . This uses up $174$ of the $994$ pairs from Step 2 and produces $116$ disjoint triples, each of sum $2985$ .