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Progression 10^(n/11)

Source: 1971 AHSME Problem 29

April 23, 2014
AMC

Problem Statement

Given the progression 10111,10211,10311,10411,…,10n1110^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}. The least positive integer nn such that the product of the first nn terms of the progression exceeds 100,000100,000 is
<spanclass=′latex−bold′>(A)</span>7<spanclass=′latex−bold′>(B)</span>8<spanclass=′latex−bold′>(C)</span>9<spanclass=′latex−bold′>(D)</span>10<spanclass=′latex−bold′>(E)</span>11<span class='latex-bold'>(A) </span>7\qquad<span class='latex-bold'>(B) </span>8\qquad<span class='latex-bold'>(C) </span>9\qquad<span class='latex-bold'>(D) </span>10\qquad <span class='latex-bold'>(E) </span>11
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