A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35 7/17. What was the number erased?	Choose the correct choice from the following answers.
A) seven 7  B) 8  C) 9  D) 15  E) 17

Rationale: "2 ) the average of the new group of numbers is 35 7 / 17 . the denominator of that fraction tells us that the total number of terms in the new list must be a multiple of 17 : 17 , 34 , 51 , 68 , 85 , etc . 3 ) since we ' re removing 1 number from a list of consecutive positive integers , we could quickly limit down the possible number of terms in the original list : if new = 17 , old = 18 , but the numbers from 1 to 18 would not have an average in the mid - 30 s ( it would be much smaller ) . eliminate this option . if new = 34 , old = 35 , but we run into the same problem . the average wo n ' t match here either . eliminate this option . if new = 51 , old = 52 , same problem here . eliminate this option . if new = 68 , old = 69 . . . . here we have a group of numbers that might just be what we ' re looking for . in theold group , the average is themiddle term = 35 , which is really close to the new average once we remove a number . now , we just have to figure out which of the 5 answers was removed ( and that changed the new average to 35 7 / 17 ) . there are a couple of ways to do this math , but i ' m going to use the answers to save myself some steps : sum of 1 to 69 , inclusive = 69 ( 35 ) = 2415 remove answer b . . . . 2415 - 8 = 2407 new average = 2407 / 68 = 35 27 / 68 now , we can compare 27 / 68 to 7 / 17 27 / 68 and 28 / 68 this is really close , but is not the answer . eliminate b . the answer is either a or c . since 28 / 68 is a bigger fraction than 27 / 68 , we need the sum ( of the 68 integers ) to be bigger . . . which means we need to remove a smaller number . final answer : a"
Answer: A