PK 2015 Q23

The sum of 1, 2, … \(n\) is \(\tfrac{1}{2}n(n + 1)\). So the smallest the sum could be if one number were deleted would be \(\tfrac{1}{2}n(n + 1) - n = \tfrac{1}{2}(n^2 + n - 2n) = \tfrac{1}{2}n(n - 1)\). Thus the mean of any \(n - 1\) of these numbers is at least \(\tfrac{1}{2}n\). So for the mean to be 4.75, we must have \(n \leqslant 9\).
Similarly, the largest the sum could be with one number deleted would be \(\tfrac{1}{2}n(n + 1) - 1 = \tfrac{1}{2}(n^2 + n - 2) = \tfrac{1}{2}(n - 1)(n + 2)\). Thus the mean of any \(n - 1\) of the numbers is at most \(\tfrac{1}{2}(n + 2)\). So we must have \(\tfrac{1}{2}(n + 2) \geqslant 4.75\) and so \(n \geqslant 8\).
Since \(4.75 = 19/4\) and the mean is obtained by dividing by \(n - 1\), we require \(n - 1\) to be a multiple of 4. So \(n = 9\). The sum of 1, … , 9 is 45 and we need the 8 numbers used to have a mean of 4.75. So their total is 38 and hence 7 is the number to be eliminated.