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> | + | 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>, 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}=5</math> This simplified to the quadratic <math>n^2+n-10=0</math> |
Revision as of 18:16, 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
, which simplified to
, or
This can be converted to
This simplified to the quadratic