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 | + | 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} | + | 68\frac{14}{17} &\leq \hspace{3mm} n &&\leq 70\frac{14}{17}, |
\end{alignat*}</cmath> | \end{alignat*}</cmath> | ||
+ | from which <math>n</math> is either <math>69</math> or <math>70.</math> | ||
+ | |||
+ | 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> | ||
+ | * 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. | ||
+ | |||
+ | * If <math>n=69,</math> then <math>(\bigstar)</math> becomes <math>2415-x=2408,</math> from which <math>x=\boxed{\textbf{(B)}\ 7}.</math> | ||
+ | |||
+ | <u><b>Remark</b></u> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~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 is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is
. What number was erased?
Solution
Suppose that there are positive integers in the set initially, so their sum is
by arithmetic series. The average of the remaining numbers is minimized when
is erased, and is maximized when
is erased.
It is clear that We write and solve a compound inequality for
from which
is either
or
Let be the number that is erased. We are given that
or
- If
then
becomes
from which
contradicting the precondition that
is a positive integer.
- If
then
becomes
from which
Remark
From note that the left side must be an integer, so must be the right side. It follows that
is divisible by
so
~MRENTHUSIASM
See Also
1982 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.