Differing views

Source: Australian Mathematical Society Gazette Puzzle Corner

Problem: An optimist and a pessimist are examining a sequence of numbers. The optimist remarks, ‘Oh jolly! The sum of any eight consecutive terms is positive!’ But the pessimist interjects, ‘Not so fast, the sum of any five consecutive terms is negative.’ Can they both be right? How long can this sequence of numbers be?

Incentive: Treat at H8 Canteen/Sodexho Cafeteria for the first person to solve it :P

Update (27/07/10): Solution: Posted in comments by Varun Jog (Berkeley Grad Student, EE IITB Alumnus) and Siddhant Agarwal (EE Senior Undergraduate, IITB)


  1. The max possible length of the sequence is 11.
    Proof that > 11 is not possible:
    Consider the first 12 numbers-- a1,a2,a3,...a12
    Since sum of 10 consecutive numbers is negative (2 blocks of 5), we can get
    a1+a2 < 0 (1)
    a3+a4 < 0 (2)
    Note also that
    a2+a3+a4 > 0 (consider the 8 block starting from a2 and cut out the last 5)
    Hence a5+a6 < 0 (3)(The 5 block from a2 has to be negative)
    Similarly a4+a5+a6 > 0 => a7+a8 < 0 (4)
    1,2,3,4 imply sum of first 8 numbers is negative, which is not true.

    A seq of length 11 could be
    10, -12, 11, 2.55, -11.8, 10, -11.8, 2.55, 11, -12, 10
    (That we can assume the seq to be symmetric simplifies the search)

  2. there is a more elegant proof which I read in problem solving strategies:


    so u see the rows sums are negative. And all column sums are positive. Contradiction! So max number possible is 11 and I am a mortal to doubt the eg of Varun Jog :)

  3. Its a famous(bad-famous) IMO problem I think.

  4. Awesome solution :) Thanks :)

    BTW, A friend at Morgan Stanley (Lets call him A - Since he does not want to be named :P) gave a better sequence
    -1.6 at positions 2, 5, 7, 10
    1 at positions 1, 3, 4, 6, 8, 9, 11


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