25 seat bar riddle

There is a bar with 25 seats in a line. The people there are anti-social so when they walk in the bar, they always try to find a seat farthest away from others. If one person walks in and find there is no seat are adjacent to nobody, that person will walk away. The bar owner wants as many people as possible. The owner can tell the first customer where to sit. all the other customers will pick the farthest possible seat from others.
So where should the first customer sit?

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SherlockHolmes Expert Asked on 29th July 2018 in Logic Puzzles.
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  • 4 Answer(s)
    Best answer

    Answer – 9 or 17

    Explanation-
    Consider a case. If the number of seats is n = (2^k)+1 for some k, then, the maximum possible seating of (n/2)+1 can be achieved by having the first person sit at position 1 or n or ceil(n/2).

    This is true since when n=(2^k)+1, then, when someone sits in the first or middle or the last seat then all the other gaps are of length in powers of 2 (as suggested above by Vinay).

    This implies that for this problem, with n=25, we need to split it into n1 and n2 such that: n1 = (2^k1)+1 and n2 = (2^k2)+1, for some k1, k2. Note that n1+n2 = n+1 = 26 (since they both share a common seat).

    Thus, the only possible values are 9 or 17.

    SherlockHolmes Expert Answered on 31st July 2018.
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    Make him sit in the 8th position

    A_23 Starter Answered on 29th July 2018.

    Ooh! not correct

    on 31st July 2018.
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    First or Last seat i think.

    Hanavi Curious Answered on 30th July 2018.

    Ooh! not correct

    on 31st July 2018.
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    I think it will make 13 people on the line

    Hanavi Curious Answered on 30th July 2018.

    Ooh! not correct

    on 31st July 2018.
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