2014 AMC 10B Problem 18

The list has total sum $11\cdot10=110$. To maximize the largest entry, minimize the sum of the other ten entries. In nondecreasing order, the sixth entry is $9$, and $8$ must be the unique mode. If $8$ appears twice, the least possible first ten entries sum to $1+2+3+8+8+9$ $+10+11+12+13$ $=77$, giving largest entry $33$. If $8$ appears three times, the least possible first ten entries are $1,1,8,8,8,9,9,10,10,11$, with sum $75$, giving largest entry $35$. If $8$ appears four or five times, the least possible sum of the first ten entries is at least $80$, so the largest entry is at most $30$. Therefore the largest possible entry is $35$, and the correct answer is E .