Question

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 \(\frac{602}{17}\). What was the number erased?

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

Hint:

If \(n\) numbers were written, the remaining \(n-1\) numbers must have a total that is an integer, so \(n-1\) is a multiple of 17. Bound \(n\) using the average.

Answer 1

The Correct Option is B

Step 1: Bound the number of integers.
Let the numbers be \(1,2,\dots,n\) and the erased number be \(x\) with \(1\le x\le n\). The average of the remaining \(n-1\) numbers is \(\frac{n(n+1)/2-x}{n-1}\). It is smallest when \(x=n\) (value \(\frac n2\)) and largest when \(x=1\) (value \(\frac{n+2}{2}\)). Hence
\[\frac n2\le \frac{602}{17}\le\frac{n+2}{2}\ \Rightarrow\ 68.8\le n\le 70.8\]so \(n=69\) or \(70\).

Step 2: Use divisibility.
The total of the remaining numbers is \((n-1)\cdot\frac{602}{17}\), an integer. Since 17 does not divide 602, \(17\mid (n-1)\). Of 69 and 70, only \(n=69\) works (\(n-1=68=4\times 17\)).

Step 3: Find \(x\).
Remaining total \(=68\times\frac{602}{17}=4\times 602=2408\). Original total \(=\frac{69\cdot 70}{2}=2415\). Erased number \(=2415-2408=7\).

Final Answer:
The correct option is (2): 7.