Question

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 $35\frac{7}{17}$ What was the number erased?

• A. 7
• B. 8
• C. 9
• D. None of these

Answer 1

The correct option is A 7

After one value is removed:

Since all of the values are Integers, the sum here must be an integer.

$Sum=\left(number\right)\times \left(average\right).$

Since the average $=35\frac{7}{17}$, 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 $37\frac{7}{17}$ will still be close to the median.

Implication:

The number of integers $=4\times 17=68$, with the result that $35\frac{7}{17}$ will be close to the median of the $68$ mostly consecutive integers.

$\therefore$ $Sum=68\times 35\frac{7}{17}=2408.$

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 $=\frac{\left(n\right)(n+1)}{2}$.

$\therefore$ $Sum=69\times \frac{70}{2}=2415.$

Removed integer = original sum - sum after one integer is removed

$=2415-2408=7.$