IMO 1977 · Problem 2

In a finite sequence of real numbers the sum of any seven successive terms is negative, and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.

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

Answer

\[ \boxed{16} \]

Solution

Suppose a sequence satisfying the conditions has at least \(17\) terms, and denote its first \(17\) terms by \(a_1,\ldots,a_{17}\). Consider \[ T=\sum_{i=1}^{11}\sum_{j=1}^{7}a_{i+j-1}. \] For each fixed \(i\), the inner sum consists of seven successive terms, so \(T<0\). But reversing the order of summation gives \[ T=\sum_{j=1}^{7}\sum_{i=1}^{11}a_{i+j-1}. \] Now each inner sum consists of eleven successive terms, so \(T>0\), a contradiction. Thus the sequence has at most \(16\) terms.

To attain this bound, take \[ 5,5,-13,5,5,5,-13,\quad 5,5,-13,5,5,5,-13,\quad 5,5. \] This is the beginning of a sequence with period seven, whose repeating block contains five entries equal to \(5\) and two equal to \(-13\). Consequently, every seven successive terms have sum \[ 5\cdot5-2\cdot13=-1<0. \]

The negative entries occur precisely at positions \(3,7,10,14\). An eleven-term block starts at one of positions \(1,\ldots,6\):

  • If it starts at \(1,2,\) or \(3\), its negative entries are at positions \(3,7,10\).
  • If it starts at \(4,5,\) or \(6\), its negative entries are at positions \(7,10,14\).

Thus every eleven successive terms contain eight entries equal to \(5\) and three equal to \(-13\), giving sum \[ 8\cdot5-3\cdot13=1>0. \] The construction satisfies both conditions, so the maximum is \(16\).

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