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

 
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Suppose that there are <math>n</math> positive integers in the set initially, so their sum is <math>\frac{n(n+1)}{2}</math> by arithmetic series. The average of the remaining numbers is minimized when <math>n</math> is erased, and is maximized when <math>1</math> is erased.
 
Suppose that there are <math>n</math> positive integers in the set initially, so their sum is <math>\frac{n(n+1)}{2}</math> by arithmetic series. The average of the remaining numbers is minimized when <math>n</math> is erased, and is maximized when <math>1</math> is erased.
  
It is clear that <math>n>1,</math> from which we write and solve a compound inequality for <math>n:</math>
+
It is clear that <math>n>1.</math> We write and solve a compound inequality for <math>n:</math>
 
<cmath>\begin{alignat*}{8}
 
<cmath>\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{\frac{n(n+1)}{2}-n}{n-1} &\leq 35\frac{7}{17} &&\leq \frac{\frac{n(n+1)}{2}-1}{n-1} \\
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\frac{n}{2} &\leq 35\frac{7}{17} &&\leq \frac{n+2}{2} \\
 
\frac{n}{2} &\leq 35\frac{7}{17} &&\leq \frac{n+2}{2} \\
 
n &\leq 70\frac{14}{17} &&\leq n+2 \\
 
n &\leq 70\frac{14}{17} &&\leq n+2 \\
68\frac{14}{17} &\leq \hspace{3mm} n &&\leq 70\frac{14}{17}.
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68\frac{14}{17} &\leq \hspace{3mm} n &&\leq 70\frac{14}{17},
 
\end{alignat*}</cmath>
 
\end{alignat*}</cmath>
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from which <math>n</math> is either <math>69</math> or <math>70.</math>
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Let <math>x</math> be the number that is erased. We are given that <math>\frac{\frac{n(n+1)}{2}-x}{n-1}=35\frac{7}{17},</math> or <cmath>\frac{n(n+1)}{2}-x=35\frac{7}{17}\cdot(n-1). \hspace{15mm}(\bigstar)</cmath>
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* If <math>n=70,</math> then <math>(\bigstar)</math> becomes <math>2485-x=\frac{41538}{17},</math> from which <math>x=\frac{707}{17},</math> contradicting the precondition that <math>x</math> is a positive integer.
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* If <math>n=69,</math> then <math>(\bigstar)</math> becomes <math>2415-x=2408,</math> from which <math>x=\boxed{\textbf{(B)}\ 7}.</math>
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<u><b>Remark</b></u>
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From <math>(\bigstar),</math> note that the left side must be an integer, so must be the right side. It follows that <math>n-1</math> is divisible by <math>17,</math> so <math>n=69.</math>
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~MRENTHUSIASM
  
 
== See Also ==
 
== See Also ==
 
{{AHSME box|year=1982|num-b=27|num-a=29}}
 
{{AHSME box|year=1982|num-b=27|num-a=29}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 04:41, 10 September 2021

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

See Also

1982 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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