23. When one number was removed from the set of positive integers from 1 to $n$, inclusive, the mean of the remaining numbers was 4.75 . What number was eliminated?
A 5
B 7
C 8
D 9
E impossible to determine
23. B The sum of $1,2, \ldots n$ is $\frac{1}{2} n(n+1)$. So the smallest the sum could be if one number were deleted would be $\frac{1}{2} n(n+1)-n=\frac{1}{2}\left(n^{2}+n-2 n\right)=\frac{1}{2} n(n-1)$. Thus the mean of any $n-1$ of these numbers is at least $\frac{1}{2} n$. 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 $\frac{1}{2} n(n+1)-1=\frac{1}{2}\left(n^{2}+n-2\right)=\frac{1}{2}(n-1)(n+2)$. Thus the mean of any $n-1$ of the numbers is at most $\frac{1}{2}(n+2)$. So we must have $\frac{1}{2}(n+2) \geqslant 4.75$ 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 $1, \ldots, 9$ 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.