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Difference between revisions of "1971 AHSME Problems/Problem 29"

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==Solution==
 
==Solution==
  
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The product of the sequence <math>10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}</math> is equal to <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}</math> since we are looking for the smallest value <math>n</math> that will create <math>100,000</math>, or <math>10^5</math>. From there, we can set up the equation <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5</math>, which simplified to <math>\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}=5</math>, or <math>1+2+3\dots n=55</math> This can be converted to <math>\frac{n(1+n)}{2}=55</math> This simplified to the quadratic <math>n^2+n-110=0</math> Or <math>(n+11)(n-10)=0</math> So <math>n=-11</math> or <math>10</math> Since only positive values of <math>n</math> work, our answer is <math>\boxed{\textbf{(D) }10}.</math>
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The product of the sequence <math>10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}</math> is equal to <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}</math> since we are looking for the smallest value <math>n</math> that will create <math>100,000</math>, or <math>10^5</math>. From there, we can set up the equation <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5</math>, which simplified to <math>\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}=5</math>, or <math>1+2+3\dots n=55</math> This can be converted to <math>\frac{n(1+n)}{2}=55</math> This simplified to the quadratic <math>n^2+n-110=0</math> Or <math>(n+11)(n-10)=0</math> So <math>n=-11</math> or <math>10</math> Since only positive values of <math>n</math> work, <math>n=10</math> makes the expression equal <math>100000</math>. However, we have to exceed <math>100000</math>, so our answer is <math>\boxed{\textbf{(E) }11}.</math>

Revision as of 14:45, 8 March 2020

Problem 29

Given the progression $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$. The least positive integer $n$ such that the product of the first $n$ terms of the progression exceeds $100,000$ is

$\textbf{(A) }7\qquad \textbf{(B) }8\qquad \textbf{(C) }9\qquad \textbf{(D) }10\qquad \textbf{(E) }11$

Solution

The product of the sequence $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ is equal to $10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}$ since we are looking for the smallest value $n$ that will create $100,000$, or $10^5$. From there, we can set up the equation $10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5$, which simplified to $\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}=5$, or $1+2+3\dots n=55$ This can be converted to $\frac{n(1+n)}{2}=55$ This simplified to the quadratic $n^2+n-110=0$ Or $(n+11)(n-10)=0$ So $n=-11$ or $10$ Since only positive values of $n$ work, $n=10$ makes the expression equal $100000$. However, we have to exceed $100000$, so our answer is $\boxed{\textbf{(E) }11}.$

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