IMO 1989 · Problem 1

Prove that the set \(\{1,2,\ldots,1989\}\) can be expressed as the disjoint union of subsets \(A_i\) \((i=1,2,\ldots,117)\) such that:

(i) Each \(A_i\) contains 17 elements;

(ii) The sum of all the elements in each \(A_i\) is the same.

Official text of the problem. Official problem paper (PDF)

Solution

We first partition \(\{1,2,\ldots,351\}\) into \(117\) triples, each having sum \(528\). Define \[ B_i=\{2i-1,\;235-i,\;294-i\} \qquad (1\le i\le59) \] and \[ C_i=\{2i,\;176-i,\;352-i\} \qquad (1\le i\le58). \] Their sums are \[ (2i-1)+(235-i)+(294-i)=528 \] and \[ 2i+(176-i)+(352-i)=528. \]

To verify that these triples form the required partition of \(\{1,\ldots,351\}\), observe that:

  • Their first entries are respectively the odd and even numbers in \(\{1,\ldots,117\}\).
  • Their second entries run through \(\{176,\ldots,234\}\) and \(\{118,\ldots,175\}\), respectively.
  • Their third entries run through \(\{235,\ldots,293\}\) and \(\{294,\ldots,351\}\), respectively.

Thus every integer from \(1\) to \(351\) occurs exactly once, and each triple contains three distinct elements.

Now partition the remaining numbers into the pairs \[ \{352+j,\;1989-j\}, \qquad j=0,1,\ldots,818. \] The first entries cover \(\{352,\ldots,1170\}\), while the second entries cover \(\{1171,\ldots,1989\}\), so these pairs are disjoint and exhaust the remaining numbers. Each pair has sum \[ 352+j+1989-j=2341. \]

There are \(819=7\cdot117\) pairs. Divide them into \(117\) batches of seven pairs and adjoin one batch to each of the \(117\) triples. Label the resulting sets \(A_1,\ldots,A_{117}\).

These sets partition \(\{1,\ldots,1989\}\). Each contains \[ 3+7\cdot2=17 \] elements, and each has sum \[ 528+7\cdot2341=16915. \] Both required properties follow. \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).