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

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== Problem 29 ==
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== Problem ==
  
 
Given the progression <math>10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}</math>.  
 
Given the progression <math>10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}</math>.  
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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, <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>
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By the rules of [[exponentiation|exponents]], the product of the first <math>n</math> terms of the sequence equals <math>10^{\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}}</math>. From here, we can set up the equation <math>10^{\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}}=100,000=10^5</math>, which simplifies to <math>\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}=5</math>, or <math>1+2+3\dots n=55</math>. By the formula for [[triangular numbers]], <math>\frac{n(n+1)}{2}=55</math>. Thus, <math>n^2+n-110=0</math>, or <math>(n+11)(n-10)=0</math>. Because <math>n>0</math>, only <math>n=10</math> makes the product equal <math>100,000</math>. However, we have to strictly exceed <math>100,000</math>, so our answer is <math>\boxed{\textbf{(E) }11}.</math>
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=28|num-a=30}}
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{{MAA Notice}}

Latest revision as of 11:14, 7 August 2024

Problem

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

By the rules of exponents, the product of the first $n$ terms of the sequence equals $10^{\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}}$. From here, we can set up the equation $10^{\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}}=100,000=10^5$, which simplifies to $\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}=5$, or $1+2+3\dots n=55$. By the formula for triangular numbers, $\frac{n(n+1)}{2}=55$. Thus, $n^2+n-110=0$, or $(n+11)(n-10)=0$. Because $n>0$, only $n=10$ makes the product equal $100,000$. However, we have to strictly exceed $100,000$, so our answer is $\boxed{\textbf{(E) }11}.$

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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