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 5 of 5: Top up every block with 7 more pairs each

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In plain words

Since every pair weighs exactly $1990$ no matter which one you pick, handing out the same number of pairs to every pan keeps the scale balanced no matter which specific pairs go where.

2985 + 7\times1990 = 16915,\qquad 3+7\times2=17

Detailed analysis

There are $994-175=819$ pairs left over from Step 2, and $819=117\times7$ . Hand out these pairs so that each of the $117$ blocks from Steps 3–4 receives exactly $7$ new, previously unused pairs. Each block grows from $3$ elements to $3+7\times2=17$ elements, and its sum grows from $2985$ to $2985+7\times1990=16915$ , the target found in Step 1 — the same value for every block, since every block received the same number of pairs of the same total value $1990$ . This gives the required partition $A_1,\ldots,A_{117}$ .