## Another brainteaser

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Edited by Richard Walker, Saturday, 27 Feb 2010, 01:17

In a cross-country run, Sven placed exactly in the middle among all participants.

Dan placed lower (i.e. did worse than Sven), in tenth place, and Lars placed sixteenth.

How many runners took part in the race?

(This is another mini-classic, I'll try to recollect where it was first published.)

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### Errm?

Let number of competitors be N, Sven's placing S and given placings D, L

S = N div 2 + 1, given.

((D = 10) and (S < D)) => (Smax = 9).

(L = 16) => (Nmin = 16) => (Smin = 16 div 2 + 1 = 9)

((Smax = 9) and (Smin = 9)) => (S = 9)

N = S * 2 - 1 = 17.

### This seems a bit terse :-)

Hi Martin

For a more general audience though?  A more verbal explanation would help I feel!

### New comment

again, I don't do maths, but this is what I have in my head.

Sven is half way, and we are not told last place. But Sven is higher than 10 (Dan's place) There is at least 16 as this is where Lars is. So half of 16 is 8, this is where Sven could be. Cannot be 10, so Sven could be at 8 or 9

He is right in the middle, so the answer has to be odd. So it cannot 19 as the middle of this is 10. So it has to be 17

### New comment

another way of looking at it:

from reduction

lars ended 16th but not last therefore more than 16 runners.

if dan was middle then there would be 21 runners

total is thus   16 > t < 21

possible solution set is thus 17, 18, 19, 20

19 and 20 can be removed as 19 is true for dan being middle, sven finished higher than dan.

solution set is thus 17, 18

we are looking for an odd number as there are an equal amount of runners before and after Sven (he is in the middle) so

t = 17

where did he place?

17 = 2n-1

n= 9