Let the numbers excluded be x-1, x and x + 1.
Then we know that
n å i=1
i - 3x = 7.5 ×(n-3) (1).
Hence:
[
n å i=4
i ]/(n-3) ³ 7.5 ³ [
n-3 å i=1
i]/(n-3).
It follows that:
n(n + 1)/2 - 6 ³ 7.5(n-3) (2)
and
7.5 ³
12
(n-2) (3) .
Hence n2 - 14n -33 = (n-3)(n-11) ³ 0 and, as we already
know n is not less than 3, this gives n ³ 11
From (2) and (3) we have 11 £ n £ 17.
From (1): n(n+1)/2 - 3x = 7.5(n - 3)
and so
x = [n(n+1)/2 - 7.5(n - 3)]/3
Using a spreadsheet to calculate x for n Î [11, 17] we find
that the solutions are: n = 11, remove 1,2, and 3 n = 15, remove 9, 10 and 11 n = 17, remove 15, 16 and 17.
See the spreadsheet
giving the solutions.