A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came and erased one number. The average of the remaining numbers is 35(7/17). What was the number erased?
A) 7
B) 8
C) 9
D) none of these
Answers
Given: A set of consecutive positive integers beginning with 1 is written on the blackboard
To find: What was the number erased?
Solution:
- Now we have given that a student came and erased one number. The average of the remaining numbers is 35(7/17).
- The sum will be an integer as all of the values are Integers.
- So the sum will be the product of number and average.
- The average is 35 + 7/17.
- The number of integers must be a multiple of 17
- Now for any evenly spaced set, average is always equal to median.
- Now the number of integers = 4 x 17 = 68
- Sum = 68 x (35 + 7/17) = 2408.
- Now for original set:
- 68 integers remain after one of the integers is removed, then the original set will contain 69 integers.
- Sum of the first n positive integers = (n)(n+1) /2
69 x 70 / 2
2415.
- So now, removed integer will be:
original sum - sum after one integer is removed
2415 - 2408 = 7
Answer:
So the erased number is 7.
Answer:
Correct option is
A
7
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
17
7
, and the sum must 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 still be evenly spaced.
As a result, the average of the remaining set 37
17
7
will still be close to the median.
Implication:
The number of integers =4×17=68, with the result that 35
17
7
will be close to the median of the 68 mostly consecutive integers.
∴ Sum=68×35
17
7
=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 =
2
(n)(n+1)
.
∴ Sum=69×
2
70
=2415.
Removed integer = original sum - sum after one integer is removed
=2415−2408=7.