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 1 of 5: Find the target sum for each block

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In plain words

Think of the $1989$ numbers as weights that must be split evenly onto $117$ pans of a scale; the arithmetic first tells us exactly how heavy each pan has to be before we even think about which numbers go where.

\sum_{i=1}^{1989} i = \frac{1989 \times 1990}{2} = 1989 \times 995, \qquad \frac{1989\times 995}{117} = 17 \times 995 = 16915

Detailed analysis

If $\{1,\ldots,1989\}$ splits into $117$ blocks of $17$ numbers with equal sums, that common sum must be the total $\frac{1989\times1990}{2}=1989\times995$ divided by $117$ , which is $17\times995=16915$ . Notice $995$ is exactly the average of $1$ and $1989$ , so a natural target is: seventeen numbers averaging to $995$ .