Let the largest of the original numbers in the set be n with S representing the sum of the deleted numbers. Also, let v=45.1 represent the average of the remaining numbers (after the deletions).
If n, n-1, n-2, n-3, and n-4 are deleted, then the mean of the remaining numbers is (n-4)/2.
If 1, 2, 3, 4, and 5 are deleted, the mean would be (n+6)/2
--> (n-4)/2 <= v <= (n+6)/2 --> 2v-6 <= n <= 2v+4 --> 85 <= n <= 94.
But (1+2+...+n - S)/(n-5)=v --> S = n(n+1)/2 - v(n-5) --> 45.1(n-5) is an integer --> n=85 only.
Now n=85 --> S=47 --> the largest possible deleted number is 37 (if the other deleted numbers are 1,2,3, and 4). |