1977 IMO Problems/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.


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]

See Also

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