After one value is removed:
Since all of the values are integers, the sum here must be an integer.
sum=(number)×(average)
Since the average
=35, and the sum be an integer, the number of integers must be a multiple of 17,
For any evenly spaced set, average = median.
After one of the consecutive integers is removed , most of the remaining set will be evenly spaced.
As a result , the average of the remaining set
37 will still be close to the median.
Implication:
The number of integers
=4×17=68, with the result that
35 will be close to the median of the 68 mostly consecutive integers.
∴ Sum
=68×35=2408 Original set:
Since 68 integers remain after one of the integers is removed, the original set contains 69 integers.
Sum of the first n positive integers
= ∴ Sum
=69=2415 Removed integer = original sum - sum after one integer is removed
=2415−2408=7.