# Difference between revisions of "1982 AHSME Problems/Problem 28"

## Problem

A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\frac{7}{17}$. What number was erased?

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{cannot be determined}$

## Solution

Suppose that there are $n$ positive integers in the set initially, so their sum is $\frac{n(n+1)}{2}$ by arithmetic series. The average of the remaining numbers is minimized when $n$ is erased, and is maximized when $1$ is erased.

It is clear that $n>1.$ We write and solve a compound inequality for $n:$ \begin{alignat*}{8} \frac{\frac{n(n+1)}{2}-n}{n-1} &\leq 35\frac{7}{17} &&\leq \frac{\frac{n(n+1)}{2}-1}{n-1} \\ \frac{n(n+1)-2n}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{n(n+1)-2}{2(n-1)} \\ \frac{n^2-n}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{n^2+n-2}{2(n-1)} \\ \frac{n(n-1)}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{(n+2)(n-1)}{2(n-1)} \\ \frac{n}{2} &\leq 35\frac{7}{17} &&\leq \frac{n+2}{2} \\ n &\leq 70\frac{14}{17} &&\leq n+2 \\ 68\frac{14}{17} &\leq \hspace{3mm} n &&\leq 70\frac{14}{17}, \end{alignat*} from which $n$ is either $69$ or $70.$

Let $x$ be the number that is erased. We are given that $\frac{\frac{n(n+1)}{2}-x}{n-1}=35\frac{7}{17},$ or $$\frac{n(n+1)}{2}-x=35\frac{7}{17}\cdot(n-1). \hspace{15mm}(\bigstar)$$

• If $n=70,$ then $(\bigstar)$ becomes $2485-x=\frac{41538}{17},$ from which $x=\frac{707}{17},$ contradicting the precondition that $x$ is a positive integer.
• If $n=69,$ then $(\bigstar)$ becomes $2415-x=2408,$ from which $x=\boxed{\textbf{(B)}\ 7}.$

Remark

From $(\bigstar),$ note that the left side must be an integer, so must be the right side. It follows that $n-1$ is divisible by $17,$ so $n=69.$

~MRENTHUSIASM