Question

A set of consecutive positive integers beginning with1 is written on the blackboard. A student came and erased one of the numbers. The average of the remaining number is \[35\dfrac{7}{{17}}\]. What was the number erased?
A.7
B.8
C.9
D.None of these

Answer 1

Hint: As the integers are added we will get an integer. Thus, from the average of the remaining numbers which is \[35\dfrac{7}{{17}}\], if we want to get the sum of the remaining integers, we have to multiply this number by a multiple 17, so as to get an integer. Thus, the final number of integers written on the blackboard will be a multiple of 17 such as 17 or 34 or 51 or 68. However we also have to make sure that the average of all the integers must be in around 35. For this we know that as the integers are equally spaced (with a difference of 1 each), the mean of the set of data will be close to the median or middle value and even after removing one of the numbers, the median or mean be very close to its initial value. Thus, the number of integers must be 68 so as to get a median around 35. After this we can find the total value of the sum of all integers before and after removing one of the integers and then subtract to get the number which is removed.

Complete step-by-step answer:
As the average of the remaining integers is \[35\dfrac{7}{{17}}\], we have to multiply the average with a multiple of 17 to get rid of the denominator and get an integer as sum of integers will also be an integer,.
We also know that as the integers are equally spaced (with a difference of 1 each), the mean of the set of data will be close to the median or middle value and even after removing one of the numbers, the median or mean will be very close to its initial value.
As we want an average of \[35\dfrac{7}{{17}}\] after removing one of the numbers, so the total number of integers after removing one of the numbers will be 68 which is the \[{4^{th}}\]multiple of 17.
Thus, initially there were 69 integers from 1 to 69. To find their sum, we can use the formula for finding the sum of first \[n\] natural numbers as the integers begin from 1 so they will also be the first 69 natural numbers.
Thus, the sum of the first 69 natural numbers will be
\[\dfrac{{n(n + 1)}}{2} \\
= \dfrac{{69(69 + 1)}}{2} \\
= \dfrac{{69(70)}}{2} \\
= 69(35) = 2415\].
Also, now there are 68 numbers left with an average of \[35\dfrac{7}{{17}}\], so the sum of these 68 numbers will be equal to the product of 68 and their average.
Also, \[35\dfrac{7}{{17}}\] can be written as \[35 + \dfrac{7}{{17}}\].
Now on multiplying 68 with their average of the remaining numbers, we get
\[68\left( {35 + \dfrac{7}{{17}}} \right) \\
= 68(35) + \dfrac{{68(7)}}{{17}} \\
= 2380 + 4(7) \\
= 2380 + 28 \\
= 2408\].
Thus, we have the sum of all the integers before removing one of the integers and their difference is \[2415 - 2408 = 7\].
Thus, 7 is the integer which was erased from the blackboard.
Hence, option (A) is the correct option.

Note: The most common mistake here is that rather than considering the total number of integers to be a multiple of 17, we directly consider it as 17 itself. It is important to approach step-by-step in this question, as first we have to find the number of integers, then the sum of integers after removing one of the integers, then the sum of integers before the removal of an integer and then finally the difference. There are a lot of calculations involved which increase the chances of making mistakes.