1977 IMO Problems/Problem 2

Problem

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.

Solution

Let $x_1,x_2,\ldots$ be the given sequence and let $s_n=x_1+x_2+\ldots+x_n$ . The conditions from the hypothesis can be now written as $s_{n+7}<s_n$ and $s_{n+11}>s_n$ for all $n\ge 1$ . We then have: $0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6<s_{17}<s_{10}<s_3<s_{14}<s_7<0,$ a contradiction. Therefore, the sequence cannot have $17$ terms. In order to show that $16$ is the answer, just take 16 real numbers satisfying $s_{10}<s_3<s_{14}<s_7<0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6$ . We have $x_1=s_1$ and $x_n=s_n-s_{n-1}$ for $n\ge 2$ . Thus we found all sequences with the given properties.

The above solution was posted and copyrighted by enescu. The original thread for this problem can be found here: [1]

1977 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1977 IMO Problems/Problem 2

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