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

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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>, we set up the equation <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5</math> |

## Revision as of 18:12, 22 August 2019

## Problem 29

Given the progression . The least positive integer such that the product of the first terms of the progression exceeds is

## Solution

The product of the sequence is equal to since we are looking for the smallest value that will create , or , we set up the equation