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

(Created page with "== Problem 29 == 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>. The least posi...")
 
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[[1971 AHSME Problems/Problem 29|Solution]]
 
  
 
==Solution==
 
==Solution==

Revision as of 17:02, 22 August 2019

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

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